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• Large Scale VHDL techniques

  - by oxinabox.ucc.asn.au

  I'm thinking about implimenting a 16 bit CPU in VHDL. A simplish CPU. ADD, MULS, NEG, BitShift, JUMP, Relitive Jump, BREQ, Relitive BREQ, i don't know somethign along these lines Probably all only working with 16bit operands. I might even cut it down and use only a single operand and a accumulator. With Some status regitsters, Carry, Zero, Neg (unless i use a Accumlator), I know how to design all the parts from logic gates, and plan to build them up from first priciples, So for my ALU I'll need to 'build' a ADDer, proably a Carry Look ahead, group adder, this adder it self is make up oa a couple of parts, wich are themselves made up of a couple of parts. Anyway, my problem is not the CPU design, or the VHDL (i know the language, more or less). It's how i should keep things organised. How should I use packages, How should I name my processes and port maps? (i've never seen the benifit of naming the port maps, or processes)

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• Can this MySQL subquery be optimised?

  - by Dan

  I have two tables, news and news_views. Every time an article is viewed, the news id, IP address and date is recorded in news_views. I'm using a query with a subquery to fetch the most viewed titles from news, by getting the total count of views in the last 24 hours for each one. It works fine except that it takes between 5-10 seconds to run, presumably because there's hundreds of thousands of rows in news_views and it has to go through the entire table before it can finish. The query is as follows, is there any way at all it can be improved? SELECT n.title , nv.views FROM news n LEFT JOIN ( SELECT news_id , count( DISTINCT ip ) AS views FROM news_views WHERE datetime >= SUBDATE(now(), INTERVAL 24 HOUR) GROUP BY news_id ) AS nv ON nv.news_id = n.id ORDER BY views DESC LIMIT 15

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• wonder about some #define tricks

  - by kingkai

  Whlie reading codes of my group project, I come across many DEFINEs, and some of them seems strange. To generalize it, please look at the following 2 examples. Example 1: #define SNPRINTF(dst, fmt, arg...) snprintf(dst, sizeof(dst), fmt, ##arg) what does "##" means in this circumstance? I've tried to delete both of them, and write codes like "char buf[1024]; SNPRINTF(buf,"%s,%s","abcd","efg");" which produced the same result. So "##" seems no use and no harm to me. Example 2: #define CLOSE(fd) do { \ if (-1 != (fd)) { \ close(fd); \ (fd) = -1; \ } \ } while (0) Necessary to stuff the inner code to the do{}while(0) statement? what's the use? Thanks!

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• Integrating Facebook and example.com

  - by user366292

  I'm creating web site where you can add idea and comment it. Last week I have been looking for Facebook integration. So you can post idea from Facebook or Example.com. You should also be able to comment ideas from Facebook and Example.com. Two interfaces, one result. Is that possible? And is the Facebook application only solution? What about Facebook group wall? -- UPDATE I have Facebook and Example.com. I would like to have full integration between those. Messages and comments sent from Facebook shows on Example.com and vice versa.

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• Can i make products on Paypal site

  - by Mirage

  I offer hosting to only 2 clients for 2 diff products. I don't want to build the site and tell them to pay. Is is possible that on my paypal account i create two products like Product1 - monthly subscription 20 product 2 - monnthly yearly subscription 150 So that i can just send them the link and it automatically gets deducted from their account. One thing more can i send same link to other client as well who has bought the same product. Can i group them in paypal system to see which one has paid and which one does not Thanks

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• In Oracle 10g, how do I see tables in schemes other than system

  - by Yasir Arsanukaev

  In the hr schema there's a table emloyees, I can query data from it specifying it implicitly SELECT count(*) FROM hr.employees while logged in as system, but when I navigate to Home-Object Browser in the browser I can't see the table employees and other tables in other schemas. Can I manage tables in schemas other than system while logged in as system? Maybe there's some setting which enables me to group tables by schema. My Oracle Database version is 10.2.0.1 Express Edition. IIRC I could see all tables in Oracle Database 10.1.x.y. Thanks.

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• Advice on e-commerce logging strategy

  - by yalestar

  I recently inherited an e-commerce app (Java/Struts) that I'm porting to Rails. The thing is, we frequently have to do forensics on orders by poring through the log files, and with the old app's logs (log4j wall of text) it's pretty hard to make sense of the individual orders when several people are placing orders simultaneously. So I'm soliciting advice on a good strategy for logging of these orders, like maybe logging each individual order to its own MongoDB collection based on unique cart ID? Or maybe group them by IP address? Something different entirely? Essentially, what is the best approach for logging of an online store so that it's easy to backtrace each user's interaction with the site?

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• MySql left join on several regs

  - by egidiocs

  Hi there! I have this table1 idproduct(PK) | date_to_go 1 2010-01-18 2 2010-02-01 3 2010-02-21 4 2010-02-03 and this other table2 that controls date_to_go updates id | idproduct(FK) | prev_date_to_go | date_to_go | update_date 1 1 2010-01-01 2010-01-05 2009-12-01 2 1 2010-01-05 2010-01-10 2009-12-20 3 1 2010-01-10 2010-01-18 2009-12-20 4 3 2010-01-20 2010-02-03 2010-01-05 So, in this example, for table1.idproduct #1 2010-01-18 is the actual date_to_go and 2010-01-01 (table2.prev_date_to_go, first reg) is the original date_to_go . using this query select v.idproduct, v.date_to_go, p.prev_date_to_go original_date_to_go from table1 v left join produto_datas p on p.idproduto = v.idproduto group by (v.idproduto) order by v.idproduto can I assume that original_date_to_go will be the first related reg of table2? idproduct | date_to_go | original_date_to_go 1 2010-01-18 2010-01-01 2 2010-02-01 NULL 3 2010-02-21 2010-01-20 4 2010-02-03 NULL

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• Ubuntu Eye-Infinity across 3 displays

  - by Peter G Mac.

  So I purchased a computer recently and have been trying to customise the display. Radeon 6800 series Ubuntu 10.10 I have three 22inch 1080P lcd monitors that are mounted together. Everything is working smooth. How do I get the 'big-desktop' display where I have one enormous display across all monitors? Linux - ATI Catalyst Control Center 11.2 does not give me an option to 'group' my profiles like the pictures on their site show with Windows. I have been searching all over for help. Much Obliged, -Peter

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• Solving a SQL Server Deadlock situation

  - by mjh41

  I am trying to find a solution that will resolve a recurring deadlock situation in SQL server. I have done some analysis on the deadlock graph generated by the profiler trace and have come up with this information: The first process (spid 58) is running this query: UPDATE cds.dbo.task_core SET nstate = 1 WHERE nmboxid = 89 AND ndrawerid = 1 AND nobjectid IN (SELECT nobjectid FROM ( SELECT nobjectid, count(nobjectid) AS counting FROM cds.dbo.task_core GROUP BY nobjectid) task_groups WHERE task_groups.counting > 1) The second process (spid 86) is running this query: INSERT INTO task_core (…) VALUES (…) spid 58 is waiting for a Shared Page lock on CDS.dbo.task_core (spid 86 holds a conflicting intent exclusive (IX) lock) spid 86 is waiting for an Intent Exclusive (IX) page lock on CDS.dbo.task_core (spid 58 holds a conflicting Update lock)

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• Execute Oracle RAC cluster commands via Solaris RBAC?

  - by David Citron

  Executing Oracle RAC cluster management commands such as $ORA_CRS_HOME/bin/crs_start requires root permissions. Using Solaris RBAC (Role-Based Access Control), one can give a non-root user permissions to execute those commands, but the commands still fail internally. Example: $pfexec /opt/11.1.0/crs/bin/crs_stop SomeArg CRS-0259: Owner of the resource does not belong to the group. Is there a complete RBAC solution for Oracle RAC or does the executor need to be root? EDIT: Note that my original /etc/security/exec_attr contained: MyProfile:suser:cmd:::/opt/11.1.0/crs/bin/crs_start:uid=0 MyProfile:suser:cmd:::/opt/11.1.0/crs/bin/crs_start.bin:uid=0 As Martin suggests below, this needed to be changed to add gid=0 as: MyProfile:suser:cmd:::/opt/11.1.0/crs/bin/crs_start:uid=0;gid=0 MyProfile:suser:cmd:::/opt/11.1.0/crs/bin/crs_start.bin:uid=0;gid=0

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• Custom Validator and specifying type of message

  - by user102533

  I have a custom validation in the enterprise validation block. The DoValidate method is as shown below. protected override void DoValidate(Double objectToValidate, object currentTarget, string key, ValidationResults validationResults) { if (!IsSalMoreThanMinWage(objectToValidate)) { //Here I need to mark this message as a "Warning" LogValidationResult(validationResults, "Salary is too low for this state", currentTarget, key); } } I'd need to mark this validation failure as a "warning" message. In the front end, when I iterate through the ValidationResults collection and grab a ValidationResult object, I would need to identify and group different types of messages and render them differently. My question is - how do I mark a failure as a warning?

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• How to get parent node in Stanford's JavaNLP?

  - by roddik

  Hello. Suppose I have such chunk of a sentence: (NP (NP (DT A) (JJ single) (NN page)) (PP (IN in) (NP (DT a) (NN wiki) (NN website)))) At a certain moment of time I have a reference to (JJ single) and I want to get the NP node binding A single page. If I get it right, that NP is the parent of the node, A and page are its siblings and it has no children (?). When I try to use the .parent() method of a tree, I always get null. The API says that's because the implementation doesn't know how to determine the parent node. Another method of interest is .ancestor(int height, Tree root), but I don't know how to get the root of the node. In both cases, since the parser knows how to indent and group trees, it must know the "parent" tree, right? How can I get it? Thanks

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• frequency of objects in an array using Ruby

  - by eastafri

  If i had a list of balls each of which has a color property. how can i cleanly get the list of balls with the most frequent color. [m1,m2,m3,m4] say, m1.color = blue m2.color = blue m3.color = red m4.color = blue [m1,m2,m4] is the list of balls with the most frequent color My Approach is to do: [m1,m2,m3,m4].group_by{|ball| ball.color}.each do |samecolor| my_items = samecolor.count end where count is defined as class Array def count k =Hash.new(0) self.each{|x|k[x]+=1} k end end my_items will be a hash of frequencies foreach same color group. My implementation could be buggy and i feel there must be a better and more smarter way. any ideas please?

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• lexical analysis gives only one output?

  - by Caffè

  I tested this example(lexe.java), but it gave me only one output. I gave this text as a reader: public class LexeTest{ private int a = 14; } And the nextToken() function is : public Category nextToken () { if (inp.findWithinHorizon (tokenPat, 0) == null) return Category.EOF; else { lastLexeme = inp.match ().group (0); if (inp.match ().start (1) != -1) return nextToken (); else if (inp.match ().start (2) != -1) return Category.IDENT; else if (inp.match ().start (3) != -1) return Category.NUMERAL; Category result = tokenMap.get (lastLexeme); if (result == null) return Category.ERROR; else return result; } } Isdie the main method: System.out.println(lexeObject.nextToken()); output is : IDENT Why? but the textfile contains multiple keywords? Anyone know what's the problem?

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• Are our PPC management agency using a robot?

  - by MQA

  We outsource the management of our Google AdWords account. Although we understand the interface ourselves, we just don't have the time to manage it, and thought we might also benefit from a pro's added experience. However, looking at the My Change History page, the only changes I see are '1 Max CPC changed' on each Ad Group - almost always at the same time every day. Are there any tools available that automate minor changes like this? Can I tell if they are ever manually logging in and making real changes??

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• Counting the most tagged tag with MySQL

  - by Jack W-H

  Hi folks My problem is that I'm trying to count which tag has been used most in a table of user-submitted code. But the problem is with the database structure. The current query I'm using is this: SELECT tag1, COUNT(tag1) AS counttag FROM code GROUP BY tag1 ORDER BY counttag DESC LIMIT 1 This is fine, except, it only counts the most often occurence of tag1 - and my database has 5 tags per post - so there's columns tag1, tag2, tag3, tag4, tag5. How do I get the highest occurring tag value from all 5 columns in one query? Jack

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• Most Rails-y way to give different views of the same resource?

  - by Nathan Long

  In Rails, is there a canonical way of giving different views of the same resource? For example, a directory of people, where each person can have multiple photos, phone numbers, email addresses, etc. The people, photos and phone numbers are actually different resources with their own RESTful actions. But when viewing people, one page might shows everyone's name and associated photos; another page is names and associated contact information, formatted for printing. Would it be more "Rails-y" to: Create additional actions on the People controller besides the RESTful ones, like "index_with_contact_info"? Create a different controller and a different group of views? Neither seems quite right to me, but the first seems more likely. Any thoughts?

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• SELECT INTO or Stored Procedure?

  - by Kerry

  Would this be better as a stored procedure or leave it as is? INSERT INTO `user_permissions` ( `user_id`, `object_id`, `type`, `view`, `add`, `edit`, `delete`, `admin`, `updated_by_user_id` ) SELECT `user_id`, $object_id, '$type', 1, 1, 1, 1, 1, $user_id FROM `user_permissions` WHERE `object_id` = $object_id_2 AND `type` = '$type_2' AND `admin` = 1 You can think of this with different objects, lets say you have groups and subgroups. If someone creates a subgroup, it is making everyone who had access to the parent group now also have access to the subgroup. I've never made a stored procedure before, but this looks like it might be time. This call be probably be called very often. Should I be creating a procedure or will the performance be insignificant?

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• What is the best way to password protect folder/page using php without a db or username

  - by Salt Packets

  What is the best way to password protect folder using php without a database or user name but using. Basically I have a page that will list contacts for organization and need to password protect that folder without having account for every user . Just one password that gets changes every so often and distributed to the group. I understand that it is not very secure but never the less I would like to know how to do this. In the best way. It would be nice if the password is remembered for a while once user entered it correctly.

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• Python regular expressions assigning to named groups

  - by None

  When you use variables (is that the correct word?) in python regular expressions like this: "blah (?P\w+)" ("value" would be the variable), how could you make the variable's value be the text after "blah " to the end of the line or to a certain character not paying any attention to the actual content of the variable. For example, this is pseudo-code for what I want: >>> import re >>> p = re.compile("say (?P<value>continue_until_text_after_assignment_is_recognized) endsay") >>> m = p.match("say Hello hi yo endsay") >>> m.group('value') 'Hello hi yo' Note: The title is probably not understandable. That is because I didn't know how to say it. Sorry if I caused any confusion.

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• Power Analysis in [R] for Two-Way Anova

  - by Thomas

  I am trying to calculate the necessary sample size for a 2x2 factorial design. I have two questions. 1) I am using the package pwr and the one way anova function to calculate the necessary sample size using the following code pwr.anova.test(k = , n = , f = , sig.level = , power = ) However, I would like to look at two way anova, since this is more efficient at estimating group means than one way anova. There is no two-way anova function that I could find. Is there a package or routine in [R] to do this? 2) Moreover, am I safe in assuming that since I am using a one-way anova power calculations, that the sample size will be more conservative (i.e. larger)?

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• CGI, python, and setgid

  - by user331398

  I'm running a compiled python cgi script (using cxfreeze) in Apache. The script, among other things, calls os.setuid(some_uid) os.setgid(some_gid) Obviously some_uid/gid are legal and I set the sticky bit for both user and group, and verified it is indeed set. However on every call i get an error os.setgid(int(self.gid)) OSError: [Errno 1] Operation not permitted As you may notice, setuid() is successful, setgid is not. Which is very weird, at least for me, though I admit I have little experience with permissions in Linux. Any thoughts/ideas are welcome. I'm using apache 2.2.15, python 2.6.5, RHEL 5.4 (kernel 2.6.18) Thank you

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• Test iPhone in-app purchases on a different bundle?

  - by Andrew Johnson

  We have a group of beta testers for iPhone app. Recently, we added in-app purchases to the app. Before this, we would send out ad hoc builds to beta testers using a separate bundle ID and name so that they could have the store build and the ad hoc build on their phones. However, it seems like we have to build the ad hoc copy with the same Bundle ID to test in-app purchases, and this means we can't send out a seperate beta copy - our beta file (annoyingly) overwrites the user's store-bought app. Is there any way to test in-app purchases in a different bundle ID? Do I need to set up fake, test in-app purchases for the test build too?

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• e2fsck extremely slow, although enough memory exists

  - by kaefert

  I've got this external USB-Disk: kaefert@blechmobil:~$ lsusb -s 2:3 Bus 002 Device 003: ID 0bc2:3320 Seagate RSS LLC As can be seen in this dmesg output, there is some problem that prevents that disk from beeing mounted: kaefert@blechmobil:~$ dmesg ... [ 113.084079] usb 2-1: new high-speed USB device number 3 using ehci_hcd [ 113.217783] usb 2-1: New USB device found, idVendor=0bc2, idProduct=3320 [ 113.217787] usb 2-1: New USB device strings: Mfr=2, Product=3, SerialNumber=1 [ 113.217790] usb 2-1: Product: Expansion Desk [ 113.217792] usb 2-1: Manufacturer: Seagate [ 113.217794] usb 2-1: SerialNumber: NA4J4N6K [ 113.435404] usbcore: registered new interface driver uas [ 113.455315] Initializing USB Mass Storage driver... [ 113.468051] scsi5 : usb-storage 2-1:1.0 [ 113.468180] usbcore: registered new interface driver usb-storage [ 113.468182] USB Mass Storage support registered. [ 114.473105] scsi 5:0:0:0: Direct-Access Seagate Expansion Desk 070B PQ: 0 ANSI: 6 [ 114.474342] sd 5:0:0:0: [sdb] 732566645 4096-byte logical blocks: (3.00 TB/2.72 TiB) [ 114.475089] sd 5:0:0:0: [sdb] Write Protect is off [ 114.475092] sd 5:0:0:0: [sdb] Mode Sense: 43 00 00 00 [ 114.475959] sd 5:0:0:0: [sdb] Write cache: enabled, read cache: enabled, doesn't support DPO or FUA [ 114.477093] sd 5:0:0:0: [sdb] 732566645 4096-byte logical blocks: (3.00 TB/2.72 TiB) [ 114.501649] sdb: sdb1 [ 114.502717] sd 5:0:0:0: [sdb] 732566645 4096-byte logical blocks: (3.00 TB/2.72 TiB) [ 114.504354] sd 5:0:0:0: [sdb] Attached SCSI disk [ 116.804408] EXT4-fs (sdb1): ext4_check_descriptors: Checksum for group 3976 failed (47397!=61519) [ 116.804413] EXT4-fs (sdb1): group descriptors corrupted! ... So I went and fired up my favorite partition manager - gparted, and told it to verify and repair the partition sdb1. This made gparted call e2fsck (version 1.42.4 (12-Jun-2012)) e2fsck -f -y -v /dev/sdb1 Although gparted called e2fsck with the "-v" option, sadly it doesn't show me the output of my e2fsck process (bugreport https://bugzilla.gnome.org/show_bug.cgi?id=467925 ) I started this whole thing on Sunday (2012-11-04_2200) evening, so about 48 hours ago, this is what htop says about it now (2012-11-06-1900): PID USER PRI NI VIRT RES SHR S CPU% MEM% TIME+ Command 3704 root 39 19 1560M 1166M 768 R 98.0 19.5 42h56:43 e2fsck -f -y -v /dev/sdb1 Now I found a few posts on the internet that discuss e2fsck running slow, for example: http://gparted-forum.surf4.info/viewtopic.php?id=13613 where they write that its a good idea to see if the disk is just that slow because maybe its damaged, and I think these outputs tell me that this is not the case in my case: kaefert@blechmobil:~$ sudo hdparm -tT /dev/sdb /dev/sdb: Timing cached reads: 3562 MB in 2.00 seconds = 1783.29 MB/sec Timing buffered disk reads: 82 MB in 3.01 seconds = 27.26 MB/sec kaefert@blechmobil:~$ sudo hdparm /dev/sdb /dev/sdb: multcount = 0 (off) readonly = 0 (off) readahead = 256 (on) geometry = 364801/255/63, sectors = 5860533160, start = 0 However, although I can read quickly from that disk, this disk speed doesn't seem to be used by e2fsck, considering tools like gkrellm or iotop or this: kaefert@blechmobil:~$ iostat -x Linux 3.2.0-2-amd64 (blechmobil) 2012-11-06 _x86_64_ (2 CPU) avg-cpu: %user %nice %system %iowait %steal %idle 14,24 47,81 14,63 0,95 0,00 22,37 Device: rrqm/s wrqm/s r/s w/s rkB/s wkB/s avgrq-sz avgqu-sz await r_await w_await svctm %util sda 0,59 8,29 2,42 5,14 43,17 160,17 53,75 0,30 39,80 8,72 54,42 3,95 2,99 sdb 137,54 5,48 9,23 0,20 587,07 22,73 129,35 0,07 7,70 7,51 16,18 2,17 2,04 Now I researched a little bit on how to find out what e2fsck is doing with all that processor time, and I found the tool strace, which gives me this: kaefert@blechmobil:~$ sudo strace -p3704 lseek(4, 41026998272, SEEK_SET) = 41026998272 write(4, "\212\354K[_\361\3nl\212\245\352\255jR\303\354\312Yv\334p\253r\217\265\3567\325\257\3766"..., 4096) = 4096 lseek(4, 48404766720, SEEK_SET) = 48404766720 read(4, "\7t\260\366\346\337\304\210\33\267j\35\377'\31f\372\252\ffU\317.y\211\360\36\240c\30`\34"..., 4096) = 4096 lseek(4, 41027002368, SEEK_SET) = 41027002368 write(4, "\232]7Ws\321\352\t\1@[+5\263\334\276{\343zZx\352\21\316`1\271[\202\350R`"..., 4096) = 4096 lseek(4, 48404770816, SEEK_SET) = 48404770816 read(4, "\17\362r\230\327\25\346//\210H\v\311\3237\323K\304\306\361a\223\311\324\272?\213\tq \370\24"..., 4096) = 4096 lseek(4, 41027006464, SEEK_SET) = 41027006464 write(4, "\367yy>x\216?=\324Z\305\351\376&\25\244\210\271\22\306}\276\237\370(\214\205G\262\360\257#"..., 4096) = 4096 lseek(4, 48404774912, SEEK_SET) = 48404774912 read(4, "\365\25\0\21|T\0\21}3t_\272\373\222k\r\177\303\1\201\261\221$\261B\232\3142\21U\316"..., 4096) = 4096 ^CProcess 3704 detached around 16 of these lines every second, so 4 read and 4 write operations every second, which I don't consider to be a lot.. And finally, my question: Will this process ever finish? If those numbers from fseek (48404774912) represent bytes, that would be something like 45 gigabytes, with this beeing a 3 terrabyte disk, which would give me 134 days to go, if the speed stays constant, and e2fsck scans the disk like this completly and only once. Do you have some advice for me? I have most of the data on that disk elsewhere, but I've put a lot of hours into sorting and merging it to this disk, so I would prefer to getting this disk up and running again, without formatting it anew. I don't think that the hardware is damaged since the disk is only a few months and since I can't see any I/O errors in the dmesg output. UPDATE: I just looked at the strace output again (2012-11-06_2300), now it looks like this: lseek(4, 1419860611072, SEEK_SET) = 1419860611072 read(4, "3#\f\2447\335\0\22A\355\374\276j\204'\207|\217V|\23\245[\7VP\251\242\276\207\317:"..., 4096) = 4096 lseek(4, 43018145792, SEEK_SET) = 43018145792 write(4, "]\206\231\342Y\204-2I\362\242\344\6R\205\361\324\177\265\317C\334V\324\260\334\275t=\10F."..., 4096) = 4096 lseek(4, 1419860615168, SEEK_SET) = 1419860615168 read(4, "\262\305\314Y\367\37x\326\245\226\226\320N\333$s\34\204\311\222\7\315\236\336\300TK\337\264\236\211n"..., 4096) = 4096 lseek(4, 43018149888, SEEK_SET) = 43018149888 write(4, "\271\224m\311\224\25!I\376\16;\377\0\223H\25Yd\201Y\342\r\203\271\24eG<\202{\373V"..., 4096) = 4096 lseek(4, 1419860619264, SEEK_SET) = 1419860619264 read(4, ";d\360\177\n\346\253\210\222|\250\352T\335M\33\260\320\261\7g\222P\344H?t\240\20\2548\310"..., 4096) = 4096 lseek(4, 43018153984, SEEK_SET) = 43018153984 write(4, "\360\252j\317\310\251G\227\335{\214`\341\267\31Y\202\360\v\374\307oq\3063\217Z\223\313\36D\211"..., 4096) = 4096 So the numbers in the lseek lines before the reads, like 1419860619264 are already a lot bigger, standing for 1.29 terabytes if those numbers are bytes, so it doesn't seem to be a linear progress on a big scale, maybe there are only some areas that need work, that have big gaps in between them. UPDATE2: Okey, big disappointment, the numbers are back to very small again (2012-11-07_0720) lseek(4, 52174548992, SEEK_SET) = 52174548992 read(4, "\374\312\22\\\325\215\213\23\0357U\222\246\370v^f(\312|f\212\362\343\375\373\342\4\204mU6"..., 4096) = 4096 lseek(4, 46603526144, SEEK_SET) = 46603526144 write(4, "\370\261\223\227\23?\4\4\217\264\320_Am\246CQ\313^\203U\253\274\204\277\2564n\227\177\267\343"..., 4096) = 4096 so either e2fsck goes over the data multiple times, or it just hops back and forth multiple times. Or my assumption that those numbers are bytes is wrong. UPDATE3: Since it's mentioned here http://forums.fedoraforum.org/showthread.php?t=282125&page=2 that you can testisk while e2fsck is running, i tried that, though not with a lot of success. When asking testdisk to display the data of my partition, this is what I get: TestDisk 6.13, Data Recovery Utility, November 2011 Christophe GRENIER <[email protected]> http://www.cgsecurity.org 1 P Linux 0 4 5 45600 40 8 732566272 Can't open filesystem. Filesystem seems damaged. And this is what strace currently gives me (2012-11-07_1030) lseek(4, 212460343296, SEEK_SET) = 212460343296 read(4, "\315Mb\265v\377Gn \24\f\205EHh\2349~\330\273\203\3375\206\10\r3=W\210\372\352"..., 4096) = 4096 lseek(4, 47347830784, SEEK_SET) = 47347830784 write(4, "]\204\223\300I\357\4\26\33+\243\312G\230\250\371*m2U\t_\215\265J \252\342Pm\360D"..., 4096) = 4096 (times are in CET)

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