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  • How to access previous VHD versions of system backup?

    - by feklee
    Quote from the 31 Oct 2009 TechNet article "Learn more about system image backup": During the first backup, the backup engine scans the source drive and copies only blocks that contain data into a .vhd file stored on the target, creating a compact view of the source drive. The next time a system image is created, only new and changed data is written to the .vhd file, and old data on the same block is moved out of the VHD and into the shadow copy storage area. Volume Shadow Copy Service is used to compute the changed data between backups, as well as to handle the process of moving the old data out to the shadow copy area on the target. This approach makes the backup fast (since only changed blocks are backed up) and efficient (since data is stored in a compact manner). When restoring the image, blocks will be restored to their original locations on the source disk. If you want to restore from an older backup, the engine reads from the shadow copy area and restores the appropriate blocks. For the last days, a daily system backup of drive C: to drive E: has been scheduled and run by Windows 7 Backup and Restore. Drive C: currently holds 233 GB of data, which fits comfortably on drive E:, a 1 TB drive, with 727 GB of free space remaining. How do I access the previous version of a VHD? I right clicked on files and folders in E:\WindowsImageBackup, and I looked for Previous Versions but always: There are no previous versions available

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  • ASP.NET MVC 3: Razor’s @: and <text> syntax

    - by ScottGu
    This is another in a series of posts I’m doing that cover some of the new ASP.NET MVC 3 features: New @model keyword in Razor (Oct 19th) Layouts with Razor (Oct 22nd) Server-Side Comments with Razor (Nov 12th) Razor’s @: and <text> syntax (today) In today’s post I’m going to discuss two useful syntactical features of the new Razor view-engine – the @: and <text> syntax support. Fluid Coding with Razor ASP.NET MVC 3 ships with a new view-engine option called “Razor” (in addition to the existing .aspx view engine).  You can learn more about Razor, why we are introducing it, and the syntax it supports from my Introducing Razor blog post.  Razor minimizes the number of characters and keystrokes required when writing a view template, and enables a fast, fluid coding workflow. Unlike most template syntaxes, you do not need to interrupt your coding to explicitly denote the start and end of server blocks within your HTML. The Razor parser is smart enough to infer this from your code. This enables a compact and expressive syntax which is clean, fast and fun to type. For example, the Razor snippet below can be used to iterate a list of products: When run, it generates output like:   One of the techniques that Razor uses to implicitly identify when a code block ends is to look for tag/element content to denote the beginning of a content region.  For example, in the code snippet above Razor automatically treated the inner <li></li> block within our foreach loop as an HTML content block because it saw the opening <li> tag sequence and knew that it couldn’t be valid C#.  This particular technique – using tags to identify content blocks within code – is one of the key ingredients that makes Razor so clean and productive with scenarios involving HTML creation. Using @: to explicitly indicate the start of content Not all content container blocks start with a tag element tag, though, and there are scenarios where the Razor parser can’t implicitly detect a content block. Razor addresses this by enabling you to explicitly indicate the beginning of a line of content by using the @: character sequence within a code block.  The @: sequence indicates that the line of content that follows should be treated as a content block: As a more practical example, the below snippet demonstrates how we could output a “(Out of Stock!)” message next to our product name if the product is out of stock: Because I am not wrapping the (Out of Stock!) message in an HTML tag element, Razor can’t implicitly determine that the content within the @if block is the start of a content block.  We are using the @: character sequence to explicitly indicate that this line within our code block should be treated as content. Using Code Nuggets within @: content blocks In addition to outputting static content, you can also have code nuggets embedded within a content block that is initiated using a @: character sequence.  For example, we have two @: sequences in the code snippet below: Notice how within the second @: sequence we are emitting the number of units left within the content block (e.g. - “(Only 3 left!”). We are doing this by embedding a @p.UnitsInStock code nugget within the line of content. Multiple Lines of Content Razor makes it easy to have multiple lines of content wrapped in an HTML element.  For example, below the inner content of our @if container is wrapped in an HTML <p> element – which will cause Razor to treat it as content: For scenarios where the multiple lines of content are not wrapped by an outer HTML element, you can use multiple @: sequences: Alternatively, Razor also allows you to use a <text> element to explicitly identify content: The <text> tag is an element that is treated specially by Razor. It causes Razor to interpret the inner contents of the <text> block as content, and to not render the containing <text> tag element (meaning only the inner contents of the <text> element will be rendered – the tag itself will not).  This makes it convenient when you want to render multi-line content blocks that are not wrapped by an HTML element.  The <text> element can also optionally be used to denote single-lines of content, if you prefer it to the more concise @: sequence: The above code will render the same output as the @: version we looked at earlier.  Razor will automatically omit the <text> wrapping element from the output and just render the content within it.  Summary Razor enables a clean and concise templating syntax that enables a very fluid coding workflow.  Razor’s smart detection of <tag> elements to identify the beginning of content regions is one of the reasons that the Razor approach works so well with HTML generation scenarios, and it enables you to avoid having to explicitly mark the beginning/ending of content regions in about 95% of if/else and foreach scenarios. Razor’s @: and <text> syntax can then be used for scenarios where you want to avoid using an HTML element within a code container block, and need to more explicitly denote a content region. Hope this helps, Scott P.S. In addition to blogging, I am also now using Twitter for quick updates and to share links. Follow me at: twitter.com/scottgu

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  • Dual Boot Oracle Solaris 11/11 and Linux (Ubuntu 11.10/grub2)

    - by HartmutStreppel
    After having worked with Open Solaris on my laptop first, then with an upgrade to Oracle Solaris 11 Express, I finally did a fresh install of Oracle Solaris 11/11, when it became available. I am not a big fan of upgrades as I know that I am not the perfect administrator and my system gets spoiled with unclean configurations, outdated packages and wrong settings that cannot be reversed. So I prefer to start from scratch. Especially with Oracle Solaris 11 I wanted to have a system just like a customer would have it in production. The installation was smooth - more or less, if I had only read the documentation a bit better in advance. For a number of reasons I prefer a dual boot system. The most important one is, that especially with mobile devices you often run into network problems. And you have a hard time figuring out where the problem is: in your laptop hardware, in the OS you are running, or really within the network. If you have an alternate OS to boot, you can exclude the OS and your hardware. This makes you feel better. The second OS should be a Linux variant - and for some not so obvious reason I decided to go with the latest Ubuntu release (11.10). It replaced a very old Open Suse installation that had not been booted for a while. I knew that it was probably best to install Ubuntu first and then Oracle Solaris 11, as this would put the right boot information for Oracle Solaris  into the MBR and onto the root partition. But then, how to enable dual boot with the 2 OSes. Searching the web one mainly finds information about dual boot of: Linux and Linux Linux and Windows I do not want to explain which wrong configurations I worked through, but I prefer to explain the final setup, which is extremely simple, and I am wondering why this is not covered as the easiest solution for most dual boot setups. I use chainloader from and to both OS'es, with the only disadvantage that I have to confirm two grub menus each time I want to boot the "other" OS. Still there were some hurdles to jump over: Ubuntu did not like getting its boot blocks being placed on the partition instead of the disk; I must admit that I do not fully understand why. But using the --force option you could get that done Ubuntu needs an active partition; that was easy to achieve grub2 uses a different numbering scheme for the partitions. That is in the docs, if you read them. BTW: The usual disclaimer is valid. There is  no guarantee that what I describe works or works well. Please back up your data carefully before trying any of this. So, Oracle Solaris 11 is installed on the first partition and Ubuntu on the third. With Ubtuntu things initially were a bit more complicated, as I did not know how to boot it. And the live CD did not offer the capability to boot the on-disk image (at least I did not find it). So I booted the live CD, mounted the Ubuntu installation at /mnt and wrote the boot blocks into the partition. This is something that does not seem to be recommended, at least grub-install refrained from doing what I intended. After a bit more research I was bold enough to use the --force option and wrote the boot blocks to /dev/sda3 using grub-install --boot-directory=/mnt/boot --force --no-floppy /dev/sda3 So, I now had a system with the Solaris boot loader in the MBR, Solaris specific boot blocks on the Solaris root partition and Ubuntu specific boot blocks in the Ubuntu partition. I just had to chain them together and I was done. Oracle Solaris 11: I have added the following lines to /rpool/boot/grub/menu.lst (be aware of the /rpool!!!!) title Ubuntu 11.10root (hd0,2)makeactivechainloader +1boot The Ubuntu root file system sits on the third partition (/dev/sda3). Ubuntu: I have added the following lines to /etc/grub.d/40_custom: menuentry "Solaris 11/11" {      set root=(hd0,1)      chainloader +1} Two things need to be mentioned: a) grub2 starts numbering partitions with 1; so my /dev/sda1 is partition 1. b) Oracle Solaris boots without the partition being made active (btw: the command to make a partition active with grub2 is "parttool (hd0,1) boot+", which currently does not work for me). As debugging grub is a bit complicated, I used the grub CLI to perform some tests and also used a tool, that I found on sourceforge.net that was able to prepare a list of all boot loaders on all partitions. This told me that the basic setup was correct. Unfortunately I lost it in the live CD environment. I hope this is helpful for some of the readers.Hartmut

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  • Advice on learning programming languages and math.

    - by Joris Ooms
    I feel like I'm getting stuck lately when it comes to learning about programming-related things; I thought I'd ask a question here and write it all down in the hope to get some pointers/advice from people. Perhaps writing it down helps me put things in perspective for myself aswell. I study Interactive Multimedia Design. This course is based on two things: graphic design on one hand, and web development on the other hand. I have quite a decent knowledge of web-related languages (the usual HTML/JS/PHP) and I'll be getting a course on ASP.NET next year. In my free time, I have learnt how to work with CodeIgniter, aswell as some diving into Ruby (and Rails) and basic iOS programming. In my first year of college I also did a class on Java (19/20 on the end result). This grade doesn't really mean anything though; I have the basics of OOP down but Java-wise, we learnt next to nothing. Considering the time I have been programming in, for example, PHP.. I can't say I'm bad at it. I'm definitely not good or great at it, but I'm decent. My teachers tell me I have the programming thing down. They just tell me I should keep on learning. So that's what I do, and I try to take in as much as possible; however, sometimes I'm unsure where to start and I have this tendency to always doubt myself. Now, for the 'question'. I want to get into iOS programming. I know iOS programming boils down to programming in Cocoa Touch and Objective-C. I also know Obj-C is a superset of C. I have done a class on C a couple of years ago, but I failed miserably. I got stuck at pointers and never really understood them.. Until like a month ago. I suddenly 'got' it. I have been working through a book on Objective-C for a week or so now, and I understand the basics (I'm at like.. chapter 6 or so). However, I keep running into similar problems as the ones I had when I did the C class: I suck at math. No, really. I come from a Latin-Modern Languages background in high school and I had nearly no math classes back then. I wanted to study Computer Science, but I failed there because of the miserable state of my mathematics knowledge. I can't explain why I'm suddenly talking about math here though, because it isn't directly related to programming.. yet it is. For example, the examples in the book I'm reading now are about programming a fraction-calculator. All good, I can do the programming when I get the formulas down.. but it takes me a full day or more to actually get to that point. I also find it hard to come up with ideas for myself. I made one small iOS app the other day and it's just a button / label kind of thing. When I press the button, it generates a random number. That's really all I could come up with. Can you 'learn' that? It probably comes down to creativity, but evidently, I'm not too great at being creative. Are there any sites or resources out there that provide something like a basic list of things you can program when you're just starting out? Maybe I'm focusing on too many things at once. I want to keep my HTML/CSS at a decent level, while learning PHP and CodeIgniter, while diving into Ruby on Rails and learning Objective-C and the iOS SDK at the same time. I just want to be good at something, I guess. The problem is that I can't seem to be happy with my PHP stuff. I want more, something 'harder'; that's why I decided to pick up the iOS thing. Like I said, I have the basics down of a lot of different languages. I can program something simple in Java, in C, in Objective-C as of this week.. but it ends there. Mostly because I can't come up with ideas for more complex applications, and also because I just doubt myself: 'Oh, that's too complex, I can never do that'. And then it ends there. To conclude my rant, let me basically rephrase my questions into a 'tl;dr' part. A. I want to get into iOS programming and I have basic knowledge of C/Objective-C. However, I struggle to come up with ideas of my own and implement them and I also suck at math which is something that isn't directly related to, yet often needed while programming. What can I do? B. I have an interest in a lot of different programming languages and I can't stop reading/learning. However, I don't feel like I'm good in anything. Should I perhaps focus on just one language for a year or longer, or keep taking it all in at the same time and hope I'll finally get them all down? C. Are there any resources out there that provide basic ideas of things I can program? I'm thinking about 'simple' command-line applications here to help me while studying C/Obj-C away from the whole iPhone SDK. Like I said, the examples in my book are mainly math-based (fraction calculator) and it's kinda hard. :( Thanks a lot for reading my post. I didn't plan it to be this long but oh well. Thanks in advance for any answers.

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  • Big Data – Buzz Words: What is HDFS – Day 8 of 21

    - by Pinal Dave
    In yesterday’s blog post we learned what is MapReduce. In this article we will take a quick look at one of the four most important buzz words which goes around Big Data – HDFS. What is HDFS ? HDFS stands for Hadoop Distributed File System and it is a primary storage system used by Hadoop. It provides high performance access to data across Hadoop clusters. It is usually deployed on low-cost commodity hardware. In commodity hardware deployment server failures are very common. Due to the same reason HDFS is built to have high fault tolerance. The data transfer rate between compute nodes in HDFS is very high, which leads to reduced risk of failure. HDFS creates smaller pieces of the big data and distributes it on different nodes. It also copies each smaller piece to multiple times on different nodes. Hence when any node with the data crashes the system is automatically able to use the data from a different node and continue the process. This is the key feature of the HDFS system. Architecture of HDFS The architecture of the HDFS is master/slave architecture. An HDFS cluster always consists of single NameNode. This single NameNode is a master server and it manages the file system as well regulates access to various files. In additional to NameNode there are multiple DataNodes. There is always one DataNode for each data server. In HDFS a big file is split into one or more blocks and those blocks are stored in a set of DataNodes. The primary task of the NameNode is to open, close or rename files and directory and regulate access to the file system, whereas the primary task of the DataNode is read and write to the file systems. DataNode is also responsible for the creation, deletion or replication of the data based on the instruction from NameNode. In reality, NameNode and DataNode are software designed to run on commodity machine build in Java language. Visual Representation of HDFS Architecture Let us understand how HDFS works with the help of the diagram. Client APP or HDFS Client connects to NameSpace as well as DataNode. Client App access to the DataNode is regulated by NameSpace Node. NameSpace Node allows Client App to connect to the DataNode based by allowing the connection to the DataNode directly. A big data file is divided into multiple data blocks (let us assume that those data chunks are A,B,C and D. Client App will later on write data blocks directly to the DataNode. Client App does not have to directly write to all the node. It just has to write to any one of the node and NameNode will decide on which other DataNode it will have to replicate the data. In our example Client App directly writes to DataNode 1 and detained 3. However, data chunks are automatically replicated to other nodes. All the information like in which DataNode which data block is placed is written back to NameNode. High Availability During Disaster Now as multiple DataNode have same data blocks in the case of any DataNode which faces the disaster, the entire process will continue as other DataNode will assume the role to serve the specific data block which was on the failed node. This system provides very high tolerance to disaster and provides high availability. If you notice there is only single NameNode in our architecture. If that node fails our entire Hadoop Application will stop performing as it is a single node where we store all the metadata. As this node is very critical, it is usually replicated on another clustered as well as on another data rack. Though, that replicated node is not operational in architecture, it has all the necessary data to perform the task of the NameNode in the case of the NameNode fails. The entire Hadoop architecture is built to function smoothly even there are node failures or hardware malfunction. It is built on the simple concept that data is so big it is impossible to have come up with a single piece of the hardware which can manage it properly. We need lots of commodity (cheap) hardware to manage our big data and hardware failure is part of the commodity servers. To reduce the impact of hardware failure Hadoop architecture is built to overcome the limitation of the non-functioning hardware. Tomorrow In tomorrow’s blog post we will discuss the importance of the relational database in Big Data. Reference: Pinal Dave (http://blog.sqlauthority.com) Filed under: Big Data, PostADay, SQL, SQL Authority, SQL Query, SQL Server, SQL Tips and Tricks, T SQL

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  • Deduping your redundancies

    - by nospam(at)example.com (Joerg Moellenkamp)
    Robin Harris of Storagemojo pointed to an interesting article about about deduplication and it's impact to the resiliency of your data against data corruption on ACM Queue. The problem in short: A considerable number of filesystems store important metadata at multiple locations. For example the ZFS rootblock is copied to three locations. Other filesystems have similar provisions to protect their metadata. However you can easily proof, that the rootblock pointer in the uberblock of ZFS for example is pointing to blocks with absolutely equal content in all three locatition (with zdb -uu and zdb -r). It has to be that way, because they are protected by the same checksum. A number of devices offer block level dedup, either as an option or as part of their inner workings. However when you store three identical blocks on them and the devices does block level dedup internally, the device may just deduplicated your redundant metadata to a block stored just once that is stored on the non-voilatile storage. When this block is corrupted, you have essentially three corrupted copies. Three hit with one bullet. This is indeed an interesting problem: A device doing deduplication doesn't know if a block is important or just a datablock. This is the reason why I like deduplication like it's done in ZFS. It's an integrated part and so important parts don't get deduplicated away. A disk accessed by a block level interface doesn't know anything about the importance of a block. A metadata block is nothing different to it's inner mechanism than a normal data block because there is no way to tell that this is important and that those redundancies aren't allowed to fall prey to some clever deduplication mechanism. Robin talks about this in regard of the Sandforce disk controllers who use a kind of dedup to reduce some of the nasty effects of writing data to flash, but the problem is much broader. However this is relevant whenever you are using a device with block level deduplication. It's just the point that you have to activate it for most implementation by command, whereas certain devices do this by default or by design and you don't know about it. However I'm not perfectly sure about that ? given that storage administration and server administration are often different groups with different business objectives I would ask your storage guys if they have activated dedup without telling somebody elase on their boxes in order to speak less often with the storage sales rep. The problem is even more interesting with ZFS. You may use ditto blocks to protect important data to store multiple copies of data in the pool to increase redundancy, even when your pool just consists out of one disk or just a striped set of disk. However when your device is doing dedup internally it may remove your redundancy before it hits the nonvolatile storage. You've won nothing. Just spend your disk quota on the the LUNs in the SAN and you make your disk admin happy because of the good dedup ratio However you can just fall in this specific "deduped ditto block"trap when your pool just consists out of a single device, because ZFS writes ditto blocks on different disks, when there is more than just one disk. Yet another reason why you should spend some extra-thought when putting your zpool on a single LUN, especially when the LUN is sliced and dices out of a large heap of storage devices by a storage controller. However I have one problem with the articles and their specific mention of ZFS: You can just hit by this problem when you are using the deduplicating device for the pool. However in the specifically mentioned case of SSD this isn't the usecase. Most implementations of SSD in conjunction with ZFS are hybrid storage pools and so rotating rust disk is used as pool and SSD are used as L2ARC/sZIL. And there it simply doesn't matter: When you really have to resort to the sZIL (your system went down, it doesn't matter of one block or several blocks are corrupt, you have to fail back to the last known good transaction group the device. On the other side, when a block in L2ARC is corrupt, you simply read it from the pool and in HSP implementations this is the already mentioned rust. In conjunction with ZFS this is more interesting when using a storage array, that is capable to do dedup and where you use LUNs for your pool. However as mentioned before, on those devices it's a user made decision to do so, and so it's less probable that you deduplicating your redundancies. Other filesystems lacking acapability similar to hybrid storage pools are more "haunted" by this problem of SSD using dedup-like mechanisms internally, because those filesystem really store the data on the the SSD instead of using it just as accelerating devices. However at the end Robin is correct: It's jet another point why protecting your data by creating redundancies by dispersing it several disks (by mirror or parity RAIDs) is really important. No dedup mechanism inside a device can dedup away your redundancy when you write it to a totally different and indepenent device.

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  • Is post-sudden-power-loss filesystem corruption on an SSD drive's ext3 partition "expected behavior"?

    - by Jeremy Friesner
    My company makes an embedded Debian Linux device that boots from an ext3 partition on an internal SSD drive. Because the device is an embedded "black box", it is usually shut down the rude way, by simply cutting power to the device via an external switch. This is normally okay, as ext3's journalling keeps things in order, so other than the occasional loss of part of a log file, things keep chugging along fine. However, we've recently seen a number of units where after a number of hard-power-cycles the ext3 partition starts to develop structural issues -- in particular, we run e2fsck on the ext3 partition and it finds a number of issues like those shown in the output listing at the bottom of this Question. Running e2fsck until it stops reporting errors (or reformatting the partition) clears the issues. My question is... what are the implications of seeing problems like this on an ext3/SSD system that has been subjected to lots of sudden/unexpected shutdowns? My feeling is that this might be a sign of a software or hardware problem in our system, since my understanding is that (barring a bug or hardware problem) ext3's journalling feature is supposed to prevent these sorts of filesystem-integrity errors. (Note: I understand that user-data is not journalled and so munged/missing/truncated user-files can happen; I'm specifically talking here about filesystem-metadata errors like those shown below) My co-worker, on the other hand, says that this is known/expected behavior because SSD controllers sometimes re-order write commands and that can cause the ext3 journal to get confused. In particular, he believes that even given normally functioning hardware and bug-free software, the ext3 journal only makes filesystem corruption less likely, not impossible, so we should not be surprised to see problems like this from time to time. Which of us is right? Embedded-PC-failsafe:~# ls Embedded-PC-failsafe:~# umount /mnt/unionfs Embedded-PC-failsafe:~# e2fsck /dev/sda3 e2fsck 1.41.3 (12-Oct-2008) embeddedrootwrite contains a file system with errors, check forced. Pass 1: Checking inodes, blocks, and sizes Pass 2: Checking directory structure Invalid inode number for '.' in directory inode 46948. Fix<y>? yes Directory inode 46948, block 0, offset 12: directory corrupted Salvage<y>? yes Entry 'status_2012-11-26_14h13m41.csv' in /var/log/status_logs (46956) has deleted/unused inode 47075. Clear<y>? yes Entry 'status_2012-11-26_10h42m58.csv.gz' in /var/log/status_logs (46956) has deleted/unused inode 47076. Clear<y>? yes Entry 'status_2012-11-26_11h29m41.csv.gz' in /var/log/status_logs (46956) has deleted/unused inode 47080. Clear<y>? yes Entry 'status_2012-11-26_11h42m13.csv.gz' in /var/log/status_logs (46956) has deleted/unused inode 47081. Clear<y>? yes Entry 'status_2012-11-26_12h07m17.csv.gz' in /var/log/status_logs (46956) has deleted/unused inode 47083. Clear<y>? yes Entry 'status_2012-11-26_12h14m53.csv.gz' in /var/log/status_logs (46956) has deleted/unused inode 47085. Clear<y>? yes Entry 'status_2012-11-26_15h06m49.csv' in /var/log/status_logs (46956) has deleted/unused inode 47088. Clear<y>? yes Entry 'status_2012-11-20_14h50m09.csv' in /var/log/status_logs (46956) has deleted/unused inode 47073. Clear<y>? yes Entry 'status_2012-11-20_14h55m32.csv' in /var/log/status_logs (46956) has deleted/unused inode 47074. Clear<y>? yes Entry 'status_2012-11-26_11h04m36.csv.gz' in /var/log/status_logs (46956) has deleted/unused inode 47078. Clear<y>? yes Entry 'status_2012-11-26_11h54m45.csv.gz' in /var/log/status_logs (46956) has deleted/unused inode 47082. Clear<y>? yes Entry 'status_2012-11-26_12h12m20.csv.gz' in /var/log/status_logs (46956) has deleted/unused inode 47084. Clear<y>? yes Entry 'status_2012-11-26_12h33m52.csv.gz' in /var/log/status_logs (46956) has deleted/unused inode 47086. Clear<y>? yes Entry 'status_2012-11-26_10h51m59.csv.gz' in /var/log/status_logs (46956) has deleted/unused inode 47077. Clear<y>? yes Entry 'status_2012-11-26_11h17m09.csv.gz' in /var/log/status_logs (46956) has deleted/unused inode 47079. Clear<y>? yes Entry 'status_2012-11-26_12h54m11.csv.gz' in /var/log/status_logs (46956) has deleted/unused inode 47087. Clear<y>? yes Pass 3: Checking directory connectivity '..' in /etc/network/run (46948) is <The NULL inode> (0), should be /etc/network (46953). Fix<y>? yes Couldn't fix parent of inode 46948: Couldn't find parent directory entry Pass 4: Checking reference counts Unattached inode 46945 Connect to /lost+found<y>? yes Inode 46945 ref count is 2, should be 1. Fix<y>? yes Inode 46953 ref count is 5, should be 4. Fix<y>? yes Pass 5: Checking group summary information Block bitmap differences: -(208264--208266) -(210062--210068) -(211343--211491) -(213241--213250) -(213344--213393) -213397 -(213457--213463) -(213516--213521) -(213628--213655) -(213683--213688) -(213709--213728) -(215265--215300) -(215346--215365) -(221541--221551) -(221696--221704) -227517 Fix<y>? yes Free blocks count wrong for group #6 (17247, counted=17611). Fix<y>? yes Free blocks count wrong (161691, counted=162055). Fix<y>? yes Inode bitmap differences: +(47089--47090) +47093 +47095 +(47097--47099) +(47101--47104) -(47219--47220) -47222 -47224 -47228 -47231 -(47347--47348) -47350 -47352 -47356 -47359 -(47457--47488) -47985 -47996 -(47999--48000) -48017 -(48027--48028) -(48030--48032) -48049 -(48059--48060) -(48062--48064) -48081 -(48091--48092) -(48094--48096) Fix<y>? yes Free inodes count wrong for group #6 (7608, counted=7624). Fix<y>? yes Free inodes count wrong (61919, counted=61935). Fix<y>? yes embeddedrootwrite: ***** FILE SYSTEM WAS MODIFIED ***** embeddedrootwrite: ********** WARNING: Filesystem still has errors ********** embeddedrootwrite: 657/62592 files (24.4% non-contiguous), 87882/249937 blocks Embedded-PC-failsafe:~# Embedded-PC-failsafe:~# e2fsck /dev/sda3 e2fsck 1.41.3 (12-Oct-2008) embeddedrootwrite contains a file system with errors, check forced. Pass 1: Checking inodes, blocks, and sizes Pass 2: Checking directory structure Directory entry for '.' in ... (46948) is big. Split<y>? yes Missing '..' in directory inode 46948. Fix<y>? yes Setting filetype for entry '..' in ... (46948) to 2. Pass 3: Checking directory connectivity '..' in /etc/network/run (46948) is <The NULL inode> (0), should be /etc/network (46953). Fix<y>? yes Pass 4: Checking reference counts Inode 2 ref count is 12, should be 13. Fix<y>? yes Pass 5: Checking group summary information embeddedrootwrite: ***** FILE SYSTEM WAS MODIFIED ***** embeddedrootwrite: 657/62592 files (24.4% non-contiguous), 87882/249937 blocks Embedded-PC-failsafe:~# Embedded-PC-failsafe:~# e2fsck /dev/sda3 e2fsck 1.41.3 (12-Oct-2008) embeddedrootwrite: clean, 657/62592 files, 87882/249937 blocks

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  • Strange Recurrent Excessive I/O Wait

    - by Chris
    I know quite well that I/O wait has been discussed multiple times on this site, but all the other topics seem to cover constant I/O latency, while the I/O problem we need to solve on our server occurs at irregular (short) intervals, but is ever-present with massive spikes of up to 20k ms a-wait and service times of 2 seconds. The disk affected is /dev/sdb (Seagate Barracuda, for details see below). A typical iostat -x output would at times look like this, which is an extreme sample but by no means rare: iostat (Oct 6, 2013) tps rd_sec/s wr_sec/s avgrq-sz avgqu-sz await svctm %util 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 16.00 0.00 156.00 9.75 21.89 288.12 36.00 57.60 5.50 0.00 44.00 8.00 48.79 2194.18 181.82 100.00 2.00 0.00 16.00 8.00 46.49 3397.00 500.00 100.00 4.50 0.00 40.00 8.89 43.73 5581.78 222.22 100.00 14.50 0.00 148.00 10.21 13.76 5909.24 68.97 100.00 1.50 0.00 12.00 8.00 8.57 7150.67 666.67 100.00 0.50 0.00 4.00 8.00 6.31 10168.00 2000.00 100.00 2.00 0.00 16.00 8.00 5.27 11001.00 500.00 100.00 0.50 0.00 4.00 8.00 2.96 17080.00 2000.00 100.00 34.00 0.00 1324.00 9.88 1.32 137.84 4.45 59.60 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 22.00 44.00 204.00 11.27 0.01 0.27 0.27 0.60 Let me provide you with some more information regarding the hardware. It's a Dell 1950 III box with Debian as OS where uname -a reports the following: Linux xx 2.6.32-5-amd64 #1 SMP Fri Feb 15 15:39:52 UTC 2013 x86_64 GNU/Linux The machine is a dedicated server that hosts an online game without any databases or I/O heavy applications running. The core application consumes about 0.8 of the 8 GBytes RAM, and the average CPU load is relatively low. The game itself, however, reacts rather sensitive towards I/O latency and thus our players experience massive ingame lag, which we would like to address as soon as possible. iostat: avg-cpu: %user %nice %system %iowait %steal %idle 1.77 0.01 1.05 1.59 0.00 95.58 Device: tps Blk_read/s Blk_wrtn/s Blk_read Blk_wrtn sdb 13.16 25.42 135.12 504701011 2682640656 sda 1.52 0.74 20.63 14644533 409684488 Uptime is: 19:26:26 up 229 days, 17:26, 4 users, load average: 0.36, 0.37, 0.32 Harddisk controller: 01:00.0 RAID bus controller: LSI Logic / Symbios Logic MegaRAID SAS 1078 (rev 04) Harddisks: Array 1, RAID-1, 2x Seagate Cheetah 15K.5 73 GB SAS Array 2, RAID-1, 2x Seagate ST3500620SS Barracuda ES.2 500GB 16MB 7200RPM SAS Partition information from df: Filesystem 1K-blocks Used Available Use% Mounted on /dev/sdb1 480191156 30715200 425083668 7% /home /dev/sda2 7692908 437436 6864692 6% / /dev/sda5 15377820 1398916 13197748 10% /usr /dev/sda6 39159724 19158340 18012140 52% /var Some more data samples generated with iostat -dx sdb 1 (Oct 11, 2013) Device: rrqm/s wrqm/s r/s w/s rsec/s wsec/s avgrq-sz avgqu-sz await svctm %util sdb 0.00 15.00 0.00 70.00 0.00 656.00 9.37 4.50 1.83 4.80 33.60 sdb 0.00 0.00 0.00 2.00 0.00 16.00 8.00 12.00 836.00 500.00 100.00 sdb 0.00 0.00 0.00 3.00 0.00 32.00 10.67 9.96 1990.67 333.33 100.00 sdb 0.00 0.00 0.00 4.00 0.00 40.00 10.00 6.96 3075.00 250.00 100.00 sdb 0.00 0.00 0.00 0.00 0.00 0.00 0.00 4.00 0.00 0.00 100.00 sdb 0.00 0.00 0.00 2.00 0.00 16.00 8.00 2.62 4648.00 500.00 100.00 sdb 0.00 0.00 0.00 0.00 0.00 0.00 0.00 2.00 0.00 0.00 100.00 sdb 0.00 0.00 0.00 1.00 0.00 16.00 16.00 1.69 7024.00 1000.00 100.00 sdb 0.00 74.00 0.00 124.00 0.00 1584.00 12.77 1.09 67.94 6.94 86.00 Characteristic charts generated with rrdtool can be found here: iostat plot 1, 24 min interval: http://imageshack.us/photo/my-images/600/yqm3.png/ iostat plot 2, 120 min interval: http://imageshack.us/photo/my-images/407/griw.png/ As we have a rather large cache of 5.5 GBytes, we thought it might be a good idea to test if the I/O wait spikes would perhaps be caused by cache miss events. Therefore, we did a sync and then this to flush the cache and buffers: echo 3 > /proc/sys/vm/drop_caches and directly afterwards the I/O wait and service times virtually went through the roof, and everything on the machine felt like slow motion. During the next few hours the latency recovered and everything was as before - small to medium lags in short, unpredictable intervals. Now my question is: does anybody have any idea what might cause this annoying behaviour? Is it the first indication of the disk array or the raid controller dying, or something that can be easily mended by rebooting? (At the moment we're very reluctant to do this, however, because we're afraid that the disks might not come back up again.) Any help is greatly appreciated. Thanks in advance, Chris. Edited to add: we do see one or two processes go to 'D' state in top, one of which seems to be kjournald rather frequently. If I'm not mistaken, however, this does not indicate the processes causing the latency, but rather those affected by it - correct me if I'm wrong. Does the information about uninterruptibly sleeping processes help us in any way to address the problem? @Andy Shinn requested smartctl data, here it is: smartctl -a -d megaraid,2 /dev/sdb yields: smartctl 5.40 2010-07-12 r3124 [x86_64-unknown-linux-gnu] (local build) Copyright (C) 2002-10 by Bruce Allen, http://smartmontools.sourceforge.net Device: SEAGATE ST3500620SS Version: MS05 Serial number: Device type: disk Transport protocol: SAS Local Time is: Mon Oct 14 20:37:13 2013 CEST Device supports SMART and is Enabled Temperature Warning Disabled or Not Supported SMART Health Status: OK Current Drive Temperature: 20 C Drive Trip Temperature: 68 C Elements in grown defect list: 0 Vendor (Seagate) cache information Blocks sent to initiator = 1236631092 Blocks received from initiator = 1097862364 Blocks read from cache and sent to initiator = 1383620256 Number of read and write commands whose size <= segment size = 531295338 Number of read and write commands whose size > segment size = 51986460 Vendor (Seagate/Hitachi) factory information number of hours powered up = 36556.93 number of minutes until next internal SMART test = 32 Error counter log: Errors Corrected by Total Correction Gigabytes Total ECC rereads/ errors algorithm processed uncorrected fast | delayed rewrites corrected invocations [10^9 bytes] errors read: 509271032 47 0 509271079 509271079 20981.423 0 write: 0 0 0 0 0 5022.039 0 verify: 1870931090 196 0 1870931286 1870931286 100558.708 0 Non-medium error count: 0 SMART Self-test log Num Test Status segment LifeTime LBA_first_err [SK ASC ASQ] Description number (hours) # 1 Background short Completed 16 36538 - [- - -] # 2 Background short Completed 16 36514 - [- - -] # 3 Background short Completed 16 36490 - [- - -] # 4 Background short Completed 16 36466 - [- - -] # 5 Background short Completed 16 36442 - [- - -] # 6 Background long Completed 16 36420 - [- - -] # 7 Background short Completed 16 36394 - [- - -] # 8 Background short Completed 16 36370 - [- - -] # 9 Background long Completed 16 36364 - [- - -] #10 Background short Completed 16 36361 - [- - -] #11 Background long Completed 16 2 - [- - -] #12 Background short Completed 16 0 - [- - -] Long (extended) Self Test duration: 6798 seconds [113.3 minutes] smartctl -a -d megaraid,3 /dev/sdb yields: smartctl 5.40 2010-07-12 r3124 [x86_64-unknown-linux-gnu] (local build) Copyright (C) 2002-10 by Bruce Allen, http://smartmontools.sourceforge.net Device: SEAGATE ST3500620SS Version: MS05 Serial number: Device type: disk Transport protocol: SAS Local Time is: Mon Oct 14 20:37:26 2013 CEST Device supports SMART and is Enabled Temperature Warning Disabled or Not Supported SMART Health Status: OK Current Drive Temperature: 19 C Drive Trip Temperature: 68 C Elements in grown defect list: 0 Vendor (Seagate) cache information Blocks sent to initiator = 288745640 Blocks received from initiator = 1097848399 Blocks read from cache and sent to initiator = 1304149705 Number of read and write commands whose size <= segment size = 527414694 Number of read and write commands whose size > segment size = 51986460 Vendor (Seagate/Hitachi) factory information number of hours powered up = 36596.83 number of minutes until next internal SMART test = 28 Error counter log: Errors Corrected by Total Correction Gigabytes Total ECC rereads/ errors algorithm processed uncorrected fast | delayed rewrites corrected invocations [10^9 bytes] errors read: 610862490 44 0 610862534 610862534 20470.133 0 write: 0 0 0 0 0 5022.480 0 verify: 2861227413 203 0 2861227616 2861227616 100872.443 0 Non-medium error count: 1 SMART Self-test log Num Test Status segment LifeTime LBA_first_err [SK ASC ASQ] Description number (hours) # 1 Background short Completed 16 36580 - [- - -] # 2 Background short Completed 16 36556 - [- - -] # 3 Background short Completed 16 36532 - [- - -] # 4 Background short Completed 16 36508 - [- - -] # 5 Background short Completed 16 36484 - [- - -] # 6 Background long Completed 16 36462 - [- - -] # 7 Background short Completed 16 36436 - [- - -] # 8 Background short Completed 16 36412 - [- - -] # 9 Background long Completed 16 36404 - [- - -] #10 Background short Completed 16 36401 - [- - -] #11 Background long Completed 16 2 - [- - -] #12 Background short Completed 16 0 - [- - -] Long (extended) Self Test duration: 6798 seconds [113.3 minutes]

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  • mkfs.xfs: libxfs_device_zero write failed: Input/output error

    - by Crazy_Bash
    I can't find a way to create a filesystem on one of my disks. first i'm geting the following output: [root@~]# mkfs.xfs /dev/sdb1 mkfs.xfs: /dev/sdb1 appears to contain a partition table (dos). mkfs.xfs: Use the -f option to force overwrite. after using -F flag: [root@~]# mkfs.xfs -f /dev/sdb1 meta-data=/dev/sdb1 isize=256 agcount=32, agsize=22892696 blks = sectsz=512 attr=2 data = bsize=4096 blocks=732566272, imaxpct=5 = sunit=0 swidth=0 blks naming =version 2 bsize=4096 ascii-ci=0 log =internal log bsize=4096 blocks=357698, version=2 = sectsz=512 sunit=0 blks, lazy-count=1 realtime =none extsz=4096 blocks=0, rtextents=0 **mkfs.xfs: libxfs_device_zero write failed: Input/output error** /dev/sdb: Disk /dev/sdb: 3001GB 1 1049kB 3001GB 3001GB primary Linux: Centos 6.3 Linux 1 2.6.32-279.el6.x86_64 #1 SMP Fri Jun 22 12:19:21 UTC 2012 x86_64 x86_64 x86_64 GNU/Linux what i've tried so far: recreating partition with parted rm 1

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  • What does this IIS memory dump mean? (reserved memory)

    - by Jesse
    My w3wp's are recycling every 60 seconds after using too much virtual memory. I ran the IIS Debug Diagnostic Tool to capture a memory dump before the worker process recycled; the most interesting part seems to be this: Virtual Allocation Summary Reserved memory 4.88 GBytes Committed memory 328.27 MBytes Mapped memory 17.36 MBytes Reserved block count 524 blocks Committed block count 1082 blocks Mapped block count 43 blocks So that 4.88 GBytes of reserved memory seems really big. But neither the DotNetMemoryAnalysis or the regular Memory Pressure Analyzer seems to tell me where that 4.88 GB went. How can I find out?

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  • Database implementation question?

    - by gundam
    consider a disk with a sector size of 512 bytes, 2000 tracks/surface, 50 sectors/track, 5 doubled sided platters, average seek time is 10 msec. Assume a block size of 1024-byte is selected. Assume a file that contains 100,000 records of 100-byte each is to be stored on the disk, and NONE of the reocd can be spanned 2 blocks. How many blocks are needed to store the entire file?? If the file is arranged sequentially on disk, how many surfaces are required?? Now, i have calculated that 10,000 blocks are needed to store 100,000 records. But i am not sure how to find out the answer of the surfaces required. I only calculated the capacity of track is 25KB and capacity of surface is 50,000 KB But I don't know how to calculate the number of surfaces... Could anyone help me how to get the answer? Thanks a lot!!

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  • Database implementation question? [closed]

    - by gundam
    consider a disk with a sector size of 512 bytes, 2000 tracks/surface, 50 sectors/track, 5 doubled sided platters, average seek time is 10 msec. Assume a block size of 1024-byte is selected. Assume a file that contains 100,000 records of 100-byte each is to be stored on the disk, and NONE of the reocd can be spanned 2 blocks. How many blocks are needed to store the entire file?? If the file is arranged sequentially on disk, how many surfaces are required?? Now, i have calculated that 10,000 blocks are needed to store 100,000 records. But i am not sure how to find out the answer of the surfaces required. I only calculated the capacity of track is 25KB and capacity of surface is 50,000 KB But I don't know how to calculate the number of surfaces... Could anyone help me how to get the answer? Thanks a lot!!

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  • Chaining CSS classes in IE6 - Trying to find a jQuery solution?

    - by Mike Baxter
    Right, perhaps I ask the impossible? I consider myself fairly new to Javscript and jQuery, but that being said, I have written some fairly complex code recently so I am definitely getting there... however I am now possed with a rather interesting issue at my current freelance contract. The previous web coder has taken a Grid-960 approach to the HTML and as a result has used chained classes to style many of the elements. The example below is typical of what can be found in the code: <div class='blocks four-col-1 orange highlight'>Some content</div> And in the css there will be different declarations for: (not actual css... but close enough) .blocks {margin-right:10px;} .orange {background-image:url(someimage.jpg);} .highlight {font-weight:bold;} .four-col-1 {width:300px;} and to make matters worse... this is in the CSS: .blocks.orange.highlight {background-colour:#dd00ff;} Anyone not familiar with this particular bug can read more on it here: http://www.ryanbrill.com/archives/multiple-classes-in-ie/ it is very real and very annoying. Without wanting to go into the merrits of not chaining classes (I told them this, but it is no longer feasible to change their approach... 100 hand coded pages into a 150 page website, no CMS... sigh) and without the luxury of being able to change the way these blocks are styled... can anyone advise me on the complexity and benefits between any of my below proposed approaches or possible other options that would adequately solve this problem. Potential Solution 1 Using conditional comments I am considering loading a jquery script only for IE6 that: Reads the class of all divs in a certain section of the page and pushes to an array creates empty boxes off screen with only one of the classes applied at a time Reads the applied CSS values for each box Re-applies these styles to the individual box, somehow bearing in mind the order in which they are called and overwriting conflicting instructions as required Potential Solution 2 read the class of all divs in a certain section of the page and push to an array Scan the document for links to style sheets Ajax grab the stylesheets and traverse looking for matching names to those in class array Apply styles as needed Potential Solution 3 Create an IE6 only stylesheet containing the exact style to be applied as a unique name (ie: class='blocks orange highlight' becomes class='blocks-orange-highlight') Traverse the document in IE6 and convert all spaces in class declarations to hyphens and reapply classes based on new style name Summary: Solution 1 allows the people at this company to apply any styles in the future and the script will adjust as needed. However it does not allow for the chained style to be added, only the individual style... it is also processor intensive and time consuming, but also the most likely to be converted into a plugin that could be used the world over Solution 2 is a potential nightmare to code. But again will allow for an endless number of updates without breaking Solution 3 will require someone at the companty to hardcode the new styles every time they make a change, and if they don't, IE6 will break. Ironically the site, whilst needing to conform to IE6 in a limited manner, does not need to run wihtout javascript (they've made the call... have JS or go away), so consider all jQuery and JS solutions to be 'game on'. Did I mention how much i hate IE6? Anyway... any thoughts or comments would be appreciated. I will continue to develop my own solution and if I discover one that can be turned into a jQuery plugin I will post it here in the comments. Regards, Mike.

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  • Testing for Adjacent Cells In a Multi-level Grid

    - by Steve
    I'm designing an algorithm to test whether cells on a grid are adjacent or not. The catch is that the cells are not on a flat grid. They are on a multi-level grid such as the one drawn below. Level 1 (Top Level) | - - - - - | | A | B | C | | - - - - - | | D | E | F | | - - - - - | | G | H | I | | - - - - - | Level 2 | -Block A- | -Block B- | | 1 | 2 | 3 | 1 | 2 | 3 | | - - - - - | - - - - - | | 4 | 5 | 6 | 4 | 5 | 6 | ... | - - - - - | - - - - - | | 7 | 8 | 9 | 7 | 8 | 9 | | - - - - - | - - - - - | | -Block D- | -Block E- | | 1 | 2 | 3 | 1 | 2 | 3 | | - - - - - | - - - - - | | 4 | 5 | 6 | 4 | 5 | 6 | ... | - - - - - | - - - - - | | 7 | 8 | 9 | 7 | 8 | 9 | | - - - - - | - - - - - | . . . . . . This diagram is simplified from my actual need but the concept is the same. There is a top level block with many cells within it (level 1). Each block is further subdivided into many more cells (level 2). Those cells are further subdivided into level 3, 4 and 5 for my project but let's just stick to two levels for this question. I'm receiving inputs for my function in the form of "A8, A9, B7, D3". That's a list of cell Ids where each cell Id has the format (level 1 id)(level 2 id). Let's start by comparing just 2 cells, A8 and A9. That's easy because they are in the same block. private static RelativePosition getRelativePositionInTheSameBlock(String v1, String v2) { RelativePosition relativePosition; if( v1-v2 == -1 ) { relativePosition = RelativePosition.LEFT_OF; } else if (v1-v2 == 1) { relativePosition = RelativePosition.RIGHT_OF; } else if (v1-v2 == -BLOCK_WIDTH) { relativePosition = RelativePosition.TOP_OF; } else if (v1-v2 == BLOCK_WIDTH) { relativePosition = RelativePosition.BOTTOM_OF; } else { relativePosition = RelativePosition.NOT_ADJACENT; } return relativePosition; } An A9 - B7 comparison could be done by checking if A is a multiple of BLOCK_WIDTH and whether B is (A-BLOCK_WIDTH+1). Either that or just check naively if the A/B pair is 3-1, 6-4 or 9-7 for better readability. For B7 - D3, they are not adjacent but D3 is adjacent to A9 so I can do a similar adjacency test as above. So getting away from the little details and focusing on the big picture. Is this really the best way to do it? Keeping in mind the following points: I actually have 5 levels not 2, so I could potentially get a list like "A8A1A, A8A1B, B1A2A, B1A2B". Adding a new cell to compare still requires me to compare all the other cells before it (seems like the best I could do for this step is O(n)) The cells aren't all 3x3 blocks, they're just that way for my example. They could be MxN blocks with different M and N for different levels. In my current implementation above, I have separate functions to check adjacency if the cells are in the same blocks, if they are in separate horizontally adjacent blocks or if they are in separate vertically adjacent blocks. That means I have to know the position of the two blocks at the current level before I call one of those functions for the layer below. Judging by the complexity of having to deal with mulitple functions for different edge cases at different levels and having 5 levels of nested if statements. I'm wondering if another design is more suitable. Perhaps a more recursive solution, use of other data structures, or perhaps map the entire multi-level grid to a single-level grid (my quick calculations gives me about 700,000+ atomic cell ids). Even if I go that route, mapping from multi-level to single level is a non-trivial task in itself.

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  • How to format FAT32 filesystem infected with windows virus and that is write protected

    - by explorex
    Hi, I have a pendrive with FAT32 filesystem. it is infected with virus dont know which but has autorun.inf and create exe file within folder. I tried to format it with various filesystems and even try to delete it with GParted but couldn't because it says it is write protected i can't even delete files. How to format it? user@explorerx:~$ sudo fdisk -l Disk /dev/sda: 160.0 GB, 160041885696 bytes 255 heads, 63 sectors/track, 19457 cylinders Units = cylinders of 16065 * 512 = 8225280 bytes Sector size (logical/physical): 512 bytes / 512 bytes I/O size (minimum/optimal): 512 bytes / 512 bytes Disk identifier: 0xbd04bd04 Device Boot Start End Blocks Id System /dev/sda1 * 1 498 3998720 82 Linux swap / Solaris Partition 1 does not end on cylinder boundary. /dev/sda2 499 19457 152287585+ f W95 Ext'd (LBA) /dev/sda5 5100 10198 40957686 7 HPFS/NTFS /dev/sda6 10199 14787 36861111 7 HPFS/NTFS /dev/sda7 14788 19457 37511743+ 7 HPFS/NTFS /dev/sda8 499 5099 36956160 83 Linux Partition table entries are not in disk order Disk /dev/sdc: 160.0 GB, 160041885696 bytes 255 heads, 63 sectors/track, 19457 cylinders Units = cylinders of 16065 * 512 = 8225280 bytes Sector size (logical/physical): 512 bytes / 512 bytes I/O size (minimum/optimal): 512 bytes / 512 bytes Disk identifier: 0xc13bc13b Device Boot Start End Blocks Id System /dev/sdc1 1 9729 78143488 7 HPFS/NTFS /dev/sdc2 9729 19457 78143488 7 HPFS/NTFS Disk /dev/sdb: 4194 MB, 4194304000 bytes 112 heads, 47 sectors/track, 1556 cylinders Units = cylinders of 5264 * 512 = 2695168 bytes Sector size (logical/physical): 512 bytes / 512 bytes I/O size (minimum/optimal): 512 bytes / 512 bytes Disk identifier: 0x00000000 Device Boot Start End Blocks Id System /dev/sdb1 2 1557 4091904 b W95 FAT32

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  • unable to format usb 1204 [daemon inhibited ]

    - by santosamaru
    i try to format my usb 1st time its work all data gone but i can't save any file at this usb . then i try to check is it working or broken here the report santos@santos:~$ sudo badblocks -v /dev/sdb [sudo] password for santos: Sorry, try again. [sudo] password for santos: Checking blocks 0 to 7824383 Checking for bad blocks (read-only test): 0.00% done, 0:00 elapsed. (0/0/0 errdone Pass completed, 0 bad blocks found. (0/0/0 errors) santos@santos:~$ sudo badblocks -v -w /dev/sdb [sudo] password for santos: Sorry, try again. [sudo] password for santos: /dev/sdb is apparently in use by the system; it's not safe to run badblocks! santos@santos:~$ how to format and fix this issues? i have read this link Formatting Pen Drive causes 'Daemon Is Inhibited' Error and it said like this when i try to move any items from desktop " the destination is read only also in this case i use google and find this http://ubuntuforums.org/showthread.php?t=1955353 article as same its not helped following user13509 suggestion ..

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  • Rawr Code Clone Analysis&ndash;Part 0

    - by Dylan Smith
    Code Clone Analysis is a cool new feature in Visual Studio 11 (vNext).  It analyzes all the code in your solution and attempts to identify blocks of code that are similar, and thus candidates for refactoring to eliminate the duplication.  The power lies in the fact that the blocks of code don't need to be identical for Code Clone to identify them, it will report Exact, Strong, Medium and Weak matches indicating how similar the blocks of code in question are.   People that know me know that I'm anal enthusiastic about both writing clean code, and taking old crappy code and making it suck less. So the possibilities for this feature have me pretty excited if it works well - and thats a big if that I'm hoping to explore over the next few blog posts. I'm going to grab the Rawr source code from CodePlex (a World Of Warcraft gear calculator engine program), run Code Clone Analysis against it, then go through the results one-by-one and refactor where appropriate blogging along the way.  My goals with this blog series are twofold: Evaluate and demonstrate Code Clone Analysis Provide some concrete examples of refactoring code to eliminate duplication and improve the code-base Here are the initial results:   Code Clone Analysis has found: 129 Exact Matches 201 Strong Matches 300 Medium Matches 193 Weak Matches Also indicated is that there was a total of 45,181 potentially duplicated lines of code that could be eliminated through refactoring.  Considering the entire solution only has 109,763 lines of code, if true, the duplicates lines of code number is pretty significant. In the next post we’ll start examining some of the individual results and determine if they really do indicate a potential refactoring.

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  • 12.04 LTS boot hangs at "SP5100 TCO timer: mmio address 0xfec000f0 already in use", didn't yesterday

    - by DarkIron112
    Dual-booting Windows 7 and Ubuntu 12.04 LTS. I went to reboot from Win to Ubu, and found a few interesting things. My POST screen is covered in blocks of epileptic colors until I hit GRUB, which continues when I try to boot into Ubuntu. These color blocks don't appear when I use my on-board VGA, so I'll just attribute to that. Grub dimensions are swapped (card vs onboard, probably), but, when interfacing with onboard VGA, the Grub Timeout Counter works and when using my card, it does not (see "[!!!]" below for more information) Booting into Ubuntu directly causes the error: SP5100 TCO timer: mmio address 0xfec000f0 already in use Booting into recovery mode, meanwhile, and then "resuming normal boot" gets me to the desktop without native 1440x900 resolution and graphic drivers can't tell the monitor it's looking at (I assume this is because it's not a full graphic boot, and as it says, some drivers won't run?) [!!!] When I reboot after going into recovery mode, the countdown timer works ONCE, puts me back into default ubuntu boot, and then does not work again until after another recovery-mode boot. Windows 7 can boot perfectly with no issues whatsoever from epilepsy color blocks or driver detection. This makes me wonder /why/ the POST screen can't handle my video card anymore. Amidst all the diagnostics, I opened my case and re-seated the videocard securely, ensuring it wasn't a loose connection-- But this did nothing to help me. Hardware I am running an NVidia GeForce GTX 8800 video card in a PCI slot. I have 4.8GiB memory, an AMD Athlon II Quad-core 640 Processor, on an MSI K9N6GM Series Mobo. Onboard video is an NVidia GeForce MCP61(V/S/P) card. Note: I did not have any of these problems yesterday, and I have been using Ubuntu intensively for a week, though it's been working flawlessly for months. I've recently been using it to mod my Android phone, perhaps I messed something up in the file system?

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  • SMART: DISK FAILURE IS IMMINENT (under 24 hours?)

    - by flix
    I have on my hard drive 2 OSes: Ubuntu 12.04 and Windows Vista( I keep it just because of school). Everything was OK on both OSes,but one day on Ubuntu I was getting awkward noises from my notebooks's hard drive and then everything stops and I couldn't do anything. On Windows everything was ok. Everytime I boot on Ubuntu I can get 5 minutes of normal run, without problems. After that the hard drive sounds crazy and nothing works. I could run S.M.A.R.T tests from a older Ubuntu CD (10.04) from the GUI(Disk Utility, or something like that and from terminal). From the GUI I got that the DISK FAILURE IS IMMINENT and I have ~700 bad blocks(or broken blocks, I had that test I while ago) on my HDD. From the terminal ( I don't remember if it was fsck or a SMART test command) I got that the HDD will fail in under 24 hours. Since then it passed 2-3 weeks. I've tried "badblocks" but after 10 hours it was still running and I had to stop it. Now I have to use cygwin and other alternatives for my linux apps on Windows. PLEASE HELP!!! How can I separate the bad blocks from Ubuntu so it wouldn't use them?

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  • Raid Shows Up as Multiple Drives - Can't Mount

    - by manyxcxi
    I have a single hard drive that the OS is installed on and I have Sil raid card installed with two matching 500GB hdds set up in Raid 0 and formatted- they're completely empty. For whatever reason they are showing up as /dev/sdb and /dev/sdc and not as a single hard drive. I used fdisk to format both raid drives as Linux raid auto (fd) but I cannot mount either device and dmraid doesn't seem to want to work, what step am I missing? When I installed 9.04 oh so long ago it seems like it recognized and automatically did everything that needed to be done, now I'm stuck. dmraid Output root@tripoli:~# dmraid -r /dev/sdc: sil, "sil_biaebhadcfcb", stripe, ok, 976771072 sectors, data@ 0 /dev/sdb: sil, "sil_biaebhadcfcb", stripe, ok, 976771072 sectors, data@ 0 root@tripoli:~# dmraid -ay RAID set "sil_biaebhadcfcb" already active fdisk Output root@tripoli:~# fdisk -l Disk /dev/sda: 500.1 GB, 500107862016 bytes 255 heads, 63 sectors/track, 60801 cylinders Units = cylinders of 16065 * 512 = 8225280 bytes Sector size (logical/physical): 512 bytes / 512 bytes I/O size (minimum/optimal): 512 bytes / 512 bytes Disk identifier: 0x000b9b01 Device Boot Start End Blocks Id System /dev/sda1 * 1 32 248832 83 Linux Partition 1 does not end on cylinder boundary. /dev/sda2 32 60802 488134657 5 Extended /dev/sda5 32 60802 488134656 8e Linux LVM Disk /dev/sdb: 500.1 GB, 500107862016 bytes 255 heads, 63 sectors/track, 60801 cylinders Units = cylinders of 16065 * 512 = 8225280 bytes Sector size (logical/physical): 512 bytes / 512 bytes I/O size (minimum/optimal): 512 bytes / 512 bytes Disk identifier: 0x6ead5c9a Device Boot Start End Blocks Id System /dev/sdb1 1 60801 488384001 fd Linux raid autodetect Disk /dev/sdc: 500.1 GB, 500107862016 bytes 255 heads, 63 sectors/track, 60801 cylinders Units = cylinders of 16065 * 512 = 8225280 bytes Sector size (logical/physical): 512 bytes / 512 bytes I/O size (minimum/optimal): 512 bytes / 512 bytes Disk identifier: 0xe6e2af28 Device Boot Start End Blocks Id System /dev/sdc1 1 60801 488384001 fd Linux raid autodetect

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  • Basic procedural generated content works, but how could I do the same in reverse?

    - by andrew
    My 2D world is made up of blocks. At the moment, I create a block and assign it a number between 1 and 4. The number assigned to the nth block is always the same (i.e if the player walks backwards or restarts the game.) and is generated in the function below. As shown here by this animation, the colours represent the number. function generate_data(n) math.randomseed(n) -- resets the random so that the 'random' number for n is always the same math.random() -- fixes lua random bug local no = math.random(4) --print(no, n) return no end Now I want to limit the next block's number - a block of 1 will always have a block 2 after it, while block 2 will either have a block 1,2 or 3 after it, etc. Before, all the blocks data was randomly generated, initially, and then saved. This data was then loaded and used instead of being randomly called. While working this way, I could specify what the next block would be easily and it would be saved for consistency. I have now removed this saving/loading in favour of procedural generation as I realised that save whiles would get very big after travelling. Back to the present. While travelling forward (to the right), it is easy to limit what the next blocks number will be. I can generate it at the same time as the other data. The problem is when travelling backwards (to the left) I can not think of a way to load the previous block so that it is always the same. Does anyone have any ideas on how I could sort this out?

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  • How to re-add RAID-10 dropped drive?

    - by thiesdiggity
    I have a problem that I can't seem to solve. We have a Ubuntu server setup with RAID-10 and two of the drives dropped out of the array. When I try to re-add them using the following command: mdadm --manage --re-add /dev/md2 /dev/sdc1 I get the following error message: mdadm: Cannot open /dev/sdc1: Device or resource busy When I do a "cat /proc/mdstat" I get the following: Personalities : [linear] [multipath] [raid0] [raid1] [raid6] [raid5] [raid4] [r$ md2 : active raid10 sdb1[0] sdd1[3] 1953519872 blocks 64K chunks 2 near-copies [4/2] [U__U] md1 : active raid1 sda2[0] sdc2[1] 468853696 blocks [2/2] [UU] md0 : active raid1 sda1[0] sdc1[1] 19530688 blocks [2/2] [UU] unused devices: <none> When I run "/sbin/mdadm --detail /dev/md2" I get the following: /dev/md2: Version : 00.90 Creation Time : Mon Sep 5 23:41:13 2011 Raid Level : raid10 Array Size : 1953519872 (1863.02 GiB 2000.40 GB) Used Dev Size : 976759936 (931.51 GiB 1000.20 GB) Raid Devices : 4 Total Devices : 2 Preferred Minor : 2 Persistence : Superblock is persistent Update Time : Thu Oct 25 09:25:08 2012 State : active, degraded Active Devices : 2 Working Devices : 2 Failed Devices : 0 Spare Devices : 0 Layout : near=2, far=1 Chunk Size : 64K UUID : c6d87d27:aeefcb2e:d4453e2e:0b7266cb Events : 0.6688691 Number Major Minor RaidDevice State 0 8 17 0 active sync /dev/sdb1 1 0 0 1 removed 2 0 0 2 removed 3 8 49 3 active sync /dev/sdd1 Output of df -h is: Filesystem Size Used Avail Use% Mounted on /dev/md1 441G 2.0G 416G 1% / none 32G 236K 32G 1% /dev tmpfs 32G 0 32G 0% /dev/shm none 32G 112K 32G 1% /var/run none 32G 0 32G 0% /var/lock none 32G 0 32G 0% /lib/init/rw tmpfs 64G 215M 63G 1% /mnt/vmware none 441G 2.0G 416G 1% /var/lib/ureadahead/debugfs /dev/mapper/RAID10VG-RAID10LV 1.8T 139G 1.6T 8% /mnt/RAID10 When I do a "fdisk -l" I can see all the drives needed for the RAID-10. The RAID-10 is part of the /dev/mapper, could that be the reason why the device is coming back as busy? Anyone have any suggestions on what I can try to get the drives back into the array? Any help would be greatly appreciated. Thanks!

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  • How to improve Minecraft-esque voxel world performance?

    - by SomeXnaChump
    After playing Minecraft I marveled a bit at its large worlds but at the same time I found them extremely slow to navigate, even with a quad core and meaty graphics card. Now I assume Minecraft is fairly slow because: A) It's written in Java, and as most of the spatial partitioning and memory management activities happen in there, it would naturally be slower than a native C++ version. B) It doesn't partition its world very well. I could be wrong on both assumptions; however it got me thinking about the best way to manage large voxel worlds. As it is a true 3D world, where a block can exist in any part of the world, it is basically a big 3D array [x][y][z], where each block in the world has a type (i.e BlockType.Empty = 0, BlockType.Dirt = 1 etc.) Now, I am assuming to make this sort of world perform well you would need to: A) Use a tree of some variety (oct/kd/bsp) to split all the cubes out; it seems like an oct/kd would be the better option as you can just partition on a per cube level not a per triangle level. B) Use some algorithm to work out which blocks can currently be seen, as blocks closer to the user could obfuscate the blocks behind, making it pointless to render them. C) Keep the block object themselves lightweight, so it is quick to add and remove them from the trees. I guess there is no right answer to this, but I would be interested to see peoples' opinions on the subject. How would you improve performance in a large voxel-based world?

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  • Unable to mount USBDRIVE Error creating moint point: Permission denied

    - by steve
    Whenever I plug a usb into my computer a window pops up and says Unable to mount [Name of USB] Error creating moint point: Permission denied steve@goliath:/$ uname -a Linux goliath 3.2.0-32-generic #51-Ubuntu SMP Wed Sep 26 21:33:09 UTC 2012 x86_64 x86_64 x86_64 GNU/Linux steve@goliath:/$ sudo fdisk -l WARNING: GPT (GUID Partition Table) detected on '/dev/sda'! The util fdisk doesn't support GPT. Use GNU Parted. Disk /dev/sda: 120.0 GB, 120034123776 bytes 255 heads, 63 sectors/track, 14593 cylinders, total 234441648 sectors Units = sectors of 1 * 512 = 512 bytes Sector size (logical/physical): 512 bytes / 512 bytes I/O size (minimum/optimal): 512 bytes / 512 bytes Disk identifier: 0x0f716ee1 Device Boot Start End Blocks Id System /dev/sda1 1 234441647 117220823+ ee GPT WARNING: GPT (GUID Partition Table) detected on '/dev/sdb'! The util fdisk doesn't support GPT. Use GNU Parted. Disk /dev/sdb: 1500.3 GB, 1500301910016 bytes 255 heads, 63 sectors/track, 182401 cylinders, total 2930277168 sectors Units = sectors of 1 * 512 = 512 bytes Sector size (logical/physical): 512 bytes / 512 bytes I/O size (minimum/optimal): 512 bytes / 512 bytes Disk identifier: 0x0f710ee1 Device Boot Start End Blocks Id System /dev/sdb1 1 2930277167 1465138583+ ee GPT Disk /dev/sdc: 16.0 GB, 16005464064 bytes 74 heads, 10 sectors/track, 42244 cylinders, total 31260672 sectors Units = sectors of 1 * 512 = 512 bytes Sector size (logical/physical): 512 bytes / 512 bytes I/O size (minimum/optimal): 512 bytes / 512 bytes Disk identifier: 0xc3072e18 Device Boot Start End Blocks Id System /dev/sdc1 8064 31260671 15626304 c W95 FAT32 (LBA) steve@goliath:/$ sudo mkdir /media/external mkdir: cannot create directory `/media/external': Permission denied steve@goliath:/$ sudo mkdir /media/usb0 mkdir: cannot create directory `/media/usb0': Permission denied steve@goliath:/$ sudo ls -l / | grep media drwxr-xr-x 3 root root 4096 Oct 3 22:48 media steve@goliath:/$ ls /media/ -a . .. MediaShare MediaShare is the the directory on my server that has all my movies and music. If there is any information I left out please let me know.

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  • How to determine if a 3D voxel-based room is sealed, efficiently

    - by NigelMan1010
    I've been having some issues with efficiently determining if large rooms are sealed in a voxel-based 3D rooms. I'm at a point where I have tried my hardest to solve the problem without asking for help, but not tried enough to give up, so I'm asking for help. To clarify, sealed being that there are no holes in the room. There are oxygen sealers, which check if the room is sealed, and seal depending on the oxygen input level. Right now, this is how I'm doing it: Starting at the block above the sealer tile (the vent is on the sealer's top face), recursively loop through in all 6 adjacent directions If the adjacent tile is a full, non-vacuum tile, continue through the loop If the adjacent tile is not full, or is a vacuum tile, check if it's adjacent blocks are, recursively. Each time a tile is checked, decrement a counter If the count hits zero, if the last block is adjacent to a vacuum tile, return that the area is unsealed If the count hits zero and the last block is not a vacuum tile, or the recursive loop ends (no vacuum tiles left) before the counter is zero, the area is sealed If the area is not sealed, run the loop again with some changes: Checking adjacent blocks for "breathable air" tile instead of a vacuum tile Instead of using a decrementing counter, continue until no adjacent "breathable air" tiles are found. Once loop is finished, set each checked block to a vacuum tile. Here's the code I'm using: http://pastebin.com/NimyKncC The problem: I'm running this check every 3 seconds, sometimes a sealer will have to loop through hundreds of blocks, and a large world with many oxygen sealers, these multiple recursive loops every few seconds can be very hard on the CPU. I was wondering if anyone with more experience with optimization can give me a hand, or at least point me in the right direction. Thanks a bunch.

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