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  • Bit reversal of an integer, ignoring integer size and endianness

    - by ??O?????
    Given an integer typedef: typedef unsigned int TYPE; or typedef unsigned long TYPE; I have the following code to reverse the bits of an integer: TYPE max_bit= (TYPE)-1; void reverse_int_setup() { TYPE bits= (TYPE)max_bit; while (bits <<= 1) max_bit= bits; } TYPE reverse_int(TYPE arg) { TYPE bit_setter= 1, bit_tester= max_bit, result= 0; for (result= 0; bit_tester; bit_tester>>= 1, bit_setter<<= 1) if (arg & bit_tester) result|= bit_setter; return result; } One just needs first to run reverse_int_setup(), which stores an integer with the highest bit turned on, then any call to reverse_int(arg) returns arg with its bits reversed (to be used as a key to a binary tree, taken from an increasing counter, but that's more or less irrelevant). Is there a platform-agnostic way to have in compile-time the correct value for max_int after the call to reverse_int_setup(); Otherwise, is there an algorithm you consider better/leaner than the one I have for reverse_int()? Thanks.

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  • Runtime error of TASM language help!

    - by dominoos
    .model small .stack 400h .data message db "hello. ", 0ah, 0dh, "$" firstdigit db ? seconddigit db ? thirddigit db ? number dw ? newnumber db ? anumber dw 0d bnumber dw 0d Firstn db 0ah, 0dh, "Enter first 3 digit number: ","$" secondn db 0ah, 0dh, "Enter second 3 digit number: ","$" messageB db 0ah, 0dh, "HCF of two number is: ","$" linebreaker db 0ah, 0dh, ' ', 0ah, 0dh, '$' .code Start: mov ax, @data ; establish access to the data segment mov ds, ax ; mov number, 0d mov dx, offset message ; print the string "yob choi 0648293" mov ah, 9h int 21h num: mov dx, offset Firstn ; print the string "put 1st 3 digit" mov ah, 9h int 21h ;run JMP FirstFirst ; jump to FirstFirst FirstFirst: ;first digit mov ah, 1d ;bios code for read a keystroke int 21h ;call bios, it is understood that the ascii code will be returned in al mov firstdigit, al ;may as well save a copy sub al, 30h ;Convert code to an actual integer cbw ;CONVERT BYTE TO WORD. This takes whatever number is in al and ;extends it to ax, doubling its size from 8 bits to 16 bits ;The first digit now occupies all of ax as an integer mov cx, 100d ;This is so we can calculate 100*1st digit +10*2nd digit + 3rd digit mul cx ;start to accumulate the 3 digit number in the variable imul cx ;it is understood that the other operand is ax ; the result will use both dx::ax ;dx will contain only leading zeros add anumber, ax ;save ;Second Digit mov ah, 1d ;bios code for read a keystroke int 21h ;call bios, it is understood that the ascii code will be returned in al mov seconddigit, al ;may as well save a copy sub al, 30h ;Convert code to an actual integer cbw ;CONVERT BYTE TO WORD. This takes whatever number is in al and ;extends it to ax, boubling its size from 8 bits to 16 bits ;The first digit now occupies all of ax as an integer mov cx, 10d ;continue to accumulate the 3 digit number in the variable mul cx ;it is understood that the other operand is ax, containing first digit ;the result will use both dx::ax ;dx will contain only leading zeros. add anumber, ax ;save ;third Digit mov ah, 1d ;samething as above int 21h ; mov thirddigit, al ; sub al, 30h ; cbw ; add anumber, ax ; jmp num2 ;go to checks Num2: mov dx, offset secondn ; print the string "put 2nd 3 digits" mov ah, 9h int 21h ;run JMP SecondSecond SecondSecond: ;first digit mov ah, 1d ;bios code for read a keystroke int 21h ;call bios, it is understood that the ascii code will be returned in al mov firstdigit, al ;may as well save a copy sub al, 30h ;Convert code to an actual integer cbw ;CONVERT BYTE TO WORD. This takes whatever number is in al and ;extends it to ax, doubling its size from 8 bits to 16 bits ;The first digit now occupies all of ax as an integer mov cx, 100d ;This is so we can calculate 100*1st digit +10*2nd digit + 3rd digit mul cx ;start to accumulate the 3 digit number in the variable imul cx ;it is understood that the other operand is ax ; the result will use both dx::ax ;dx will contain only leading zeros add bnumber, ax ;save ;Second Digit mov ah, 1d ;bios code for read a keystroke int 21h ;call bios, it is understood that the ascii code will be returned in al mov seconddigit, al ;may as well save a copy sub al, 30h ;Convert code to an actual integer cbw ;CONVERT BYTE TO WORD. This takes whatever number is in al and ;extends it to ax, boubling its size from 8 bits to 16 bits ;The first digit now occupies all of ax as an integer mov cx, 10d ;continue to accumulate the 3 digit number in the variable mul cx ;it is understood that the other operand is ax, containing first digit ;the result will use both dx::ax ;dx will contain only leading zeros. add bnumber, ax ;save ;third Digit mov ah, 1d ;samething as above int 21h ; mov thirddigit, al ; sub al, 30h ; cbw ; add bnumber, ax ; jmp compare ;go to compare compare: CMP ax, anumber ;comparing numbB and Number JA comp1 ;go to comp1 if anumber is bigger CMP ax, anumber ; JB comp2 ;go to comp2 if anumber is smaller CMP ax, anumber ; JE equal ;go to equal if two numbers are the same JMP compare ;go to compare (avioding error) comp1: SUB ax, anumber; subtract smaller number from bigger number JMP compare ; comp2: SUB anumber, ax; subtract smaller number from bigger number JMP compare ; equal: mov ah, 9d ;make linkbreak after the 2nd 3 digit number mov dx, offset linebreaker int 21h mov ah, 9d ;print "HCF of two number is:" mov dx, offset messageB int 21h mov ax,anumber ;copying 2nd number into ax add al,30h ; converting to ascii mov newnumber,al ; copying from low part of register into newnumb mov ah, 2d ;bios code for print a character mov dl, newnumber ;we had saved the ascii code here int 21h ;call to bios JMP exit; exit: mov ah, 4ch int 21h ;exit the program End hi, this is a program that finds highest common factor of 2 different 3digit number. if i put 200, 235,312 (low numbers) it works fine. but if i put 500, 550, 654(bigger number) the program crashes after the 2nd 3digit number is entered. can you help me to find out what problem is?

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  • what does macosx-version-min imply?

    - by Quincy
    When I pass compiler flag "-mmacosx-version-min=10.5", what does it mean? I think it implies the result binary is x86, not ppc, but is it 32 bits or 64 bits? I'm compiling on snow leopard, so default output binary is 64 bits. I'm not passing -universal, it's not 32bit-64bit universal binary, I think.

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  • Why will this code compile using ifort compiler and not when using gfortran compiler? Help!

    - by CuriousCompiler
    I'm rewriting some code to make a program compile with the gfortran compiler as opposed to ifort compiler I usually use. The code follows: _Subroutine SlideBits (WORD, BITS, ADDR) Implicit None Integer(4) WORD Integer(4) BITS Integer(4) ADDR Integer(4) ADDR1 ADDR1 = 32 - ADDR WORD = (WORD .And. (.Not.ISHFT(1,ADDR1))) .Or. ISHFT(BITS,ADDR1) End_ When I compile the above code using the gfortran compiler, I recieve this error: WORD = (WORD .And. (.Not.ISHFT(1,ADDR1))) .Or. ISHFT(BITS,ADDR1) Error: Operand of .NOT. operator at (1) is INTEGER(4) All three of the variables coming into the subroutine are integers. I've looked around a bit and the gfortran wiki states that the gfortran compiler should be able to handle logical statments being applied to integer values. Several other sites I've visited either quote from the gnu wiki or agree with it. This is the first time I've seen this error as the Intel Fortran compiler (ifort) I normally use compiles cleanly.

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  • How to get bit rotation function to accept any bit size?

    - by calccrypto
    i have these 2 functions i got from some other code def ROR(x, n): mask = (2L**n) - 1 mask_bits = x & mask return (x >> n) | (mask_bits << (32 - n)) def ROL(x, n): return ROR(x, 32 - n) and i wanted to use them in a program, where 16 bit rotations are required. however, there are also other functions that require 32 bit rotations, so i wanted to leave the 32 in the equation, so i got: def ROR(x, n, bits = 32): mask = (2L**n) - 1 mask_bits = x & mask return (x >> n) | (mask_bits << (bits - n)) def ROL(x, n, bits = 32): return ROR(x, bits - n) however, the answers came out wrong when i tested this set out. yet, the values came out correctly when the code is def ROR(x, n): mask = (2L**n) - 1 mask_bits = x & mask return (x >> n) | (mask_bits << (16 - n)) def ROL(x, n,bits): return ROR(x, 16 - n) what is going on and how do i fix this?

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  • implementing cryptographic algorithms, specifically the key expansion part

    - by masseyc
    Hey, recently I picked up a copy of Applied Cryptography by Bruce Schneier and it's been a good read. I now understand how several algorithms outlined in the book work, and I'd like to start implementing a few of them in C. One thing that many of the algorithms have in common is dividing an x-bit key, into several smaller y-bit keys. For example, blowfish's key, X, is 64-bits, but you are required to break it up into two 32-bit halves; Xl and Xr. This is where I'm getting stuck. I'm fairly decent with C, but I'm not the strongest when it comes to bitwise operators and the like. After some help on IRC, I managed to come up with these two macros: #define splitup(a, b, c) {b = a >> 32; c = a & 0xffffffff; } #define combine(a, b, c) {a = (c << 32) | a;} Where a is 64 bits and b and c are 32 bits. However, the compiler warns me about the fact that I'm shifting a 32 bit variable by 32 bits. My questions are these: what's bad about shifting a 32-bit variable 32 bits? I'm guessing it's undefined, but these macros do seem to be working. Also, would you suggest I go about this another way? As I said, I'm fairly familiar with C, but bitwise operators and the like still give me a headache.

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  • What is the proper way to declare a specialization of a template for another template type?

    - by Head Geek
    The usual definition for a specialization of a template function is something like this: class Foo { [...] }; namespace std { template<> void swap(Foo& left, Foo& right) { [...] } } // namespace std But how do you properly define the specialization when the type it's specialized on is itself a template? Here's what I've got: template <size_t Bits> class fixed { [...] }; namespace std { template<size_t Bits> void swap(fixed<Bits>& left, fixed<Bits>& right) { [...] } } // namespace std Is this the right way to declare swap? It's supposed to be a specialization of the template function std::swap, but I can't tell whether the compiler is seeing it as such, or whether it thinks that it's an overload of it or something.

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  • What platforms have something other than 8-bit char?

    - by Craig McQueen
    Every now and then, someone on SO points out that char (aka 'byte') isn't necessarily 8 bits. It seems that 8-bit char is almost universal. I would have thought that for mainstream platforms, it is necessary to have an 8-bit char to ensure its viability in the marketplace. Both now and historically, what platforms use a char that is not 8 bits, and why would they differ from the "normal" 8 bits? When writing code, and thinking about cross-platform support (e.g. for general-use libraries), what sort of consideration is it worth giving to platforms with non-8-bit char? In the past I've come across some Analog Devices DSPs for which char is 16 bits. DSPs are a bit of a niche architecture I suppose. (Then again, at the time hand-coded assembler easily beat what the available C compilers could do, so I didn't really get much experience with C on that platform.)

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  • encryption of a single character

    - by SystemicPlural
    What is the minimum number of bits needed to represent a single character of encrypted text. eg, if I wanted to encrypt the letter 'a', how many bits would I require. (assume there are many singly encrypted characters using the same key.) Am I right in thinking that it would be the size of the key. eg 256 bits?

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  • RSA key length and export limitations

    - by Alex Stamper
    I know, there are a lot of limitations to the length of used key (import and export limitations for nearly each country). Usually, it varies from 64 to 256 bits. To use more bits, it is obligatory to ask permission from authorities. But it is recommended to use 1024 bits keys for RSA as minimum! Does it mean that I cannot just use RSA without any problems with law and so on?

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  • And now for a complete change of direction from C++ function pointers

    - by David
    I am building a part of a simulator. We are building off of a legacy simulator, but going in different direction, incorporating live bits along side of the simulated bits. The piece I am working on has to, effectively route commands from the central controller to the various bits. In the legacy code, there is a const array populated with an enumerated type. A command comes in, it is looked up in the table, then shipped off to a switch statement keyed by the enumerated type. The type enumeration has a choice VALID_BUT_NOT_SIMULATED, which is effectively a no-op from the point of the sim. I need to turn those no-ops into commands to actual other things [new simulated bits| live bits]. The new stuff and the live stuff have different interfaces than the old stuff [which makes me laugh about the shill job that it took to make it all happen, but that is a topic for a different discussion]. I like the array because it is a very apt description of the live thing this chunk is simulating [latching circuits by row and column]. I thought that I would try to replace the enumerated types in the array with pointers to functions and call them directly. This would be in lieu of the lookup+switch.

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  • bit ordering and endianess

    - by Neeraj
    I am reading a file byte-by-byte. Say for example i have this byte: 0x41 (0100 0001) represented in hex. Now, I want the first three bits of this byte, i.e (010). I can use bitwise logic to extract the first three bits, but my question is will the first three bits be independent of endianess of the machine.(i.e they can't be 001)? Thanks,

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  • Help me understand why page sizes are a power of 2?

    - by eric
    Answer I need help with is: Recall that paging is implemented by breaking up an address into a page and offset number. It is most efficient to break the address into X page bits and Y offset bits, rather than perform arithmetic on the address to calculate the page number and offset. Because each bit position represents a power of 2, splitting an address between bits results in a page size that is a power of 2. i don't quite understand this answer, can anyone give a simpler explanation?

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  • Parsing a string in c#

    - by Jay
    Hi, Suppose there is an xml file like below: <Instances> <Bits = "16" XCoord = "64" YCoord = "64" ZCoord = "64" FileType="jpeg" Location="C:\Series1\Image1.jpg" ImageNumber = "1"/> <Bits = "16" XCoord = "64" YCoord = "64" ZCoord = "64" FileType="jpeg" Location="C:\Series1\Image2.jpg" ImageNumber = "2"/> <Bits = "16" XCoord = "64" YCoord = "64" ZCoord = "64" FileType="jpeg" Location="C:\Series1\Image3.jpg" ImageNumber = "3"/> <Bits = "16" XCoord = "64" YCoord = "64" ZCoord = "64" FileType="jpeg" Location="C:\Series1\Image4.jpg" ImageNumber = "4"/> <Bits = "16" XCoord = "64" YCoord = "64" ZCoord = "64" FileType="jpeg" Location="C:\Series1\Image5.jpg" ImageNumber = "5"/> </Instances> This xml file is read as a string and passed on to a function. This xml file has information about a particular image file. I want to extract the location of all the image files from this string. So whatever is value of "location" filed i need to collect all those value. What is the best way to achieve this in C#. Thanks,

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  • Why won't my PHP script work?

    - by sadicool
    I have a script that reports the following error: Warning: mysql_connect() [function.mysql-connect]: Access denied for user 'admin'@'localhost' (using password: YES) in C:\wamp\www\bits\includes\connect.inc.php on line 10 Notice: Undefined variable: l_error in C:\wamp\www\bits\includes\connect.inc.php on line 12 Notice: Undefined variable: l_cannotconnecttodatabase in C:\wamp\www\bits\includes\connect.inc.php on line 12 Why would this be?

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  • g++, R_X86_64_32S : what is it ?

    - by Skypers
    Hello. I write a 3D engine in C++ with OpenGL. I usually work on this project on my archlinux 64 bits, but on theese holidays I do on a 32 bits system. I use subversion, and since the last svn up on my 64 bits system, I've got errors : http://pastebin.be/23730 core, wrapper and interface are compilet using the -fPIC option, I do not understand so ... Thanks :)

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  • Estimating the size of a tree

    - by Full Decent
    I'd like to estimate the number of leaves in a large tree structure for which I can't visit every node exhaustively. Is this algorithm appropriate? Does it have a name? Also, please pedant if I am using any terms improperly. sum_trials = 0 num_trials = 0 WHILE time_is_not_up bits = 0 ptr = tree.root WHILE count(ptr.children) > 0 bits += log2(count(ptr.children)) ptr = ptr.children[rand()%count(ptr.children)] sum_trials += bits num_trials++ estimated_tree_size = 2^(sum_trials/num_trials)

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  • DRBD not syncing between my nodes when IP is reset

    - by ramdaz
    I am trying to setup DRBD by following the article at http://www.howtoforge.com/setting-up-network-raid1-with-drbd-on-ubuntu-11.10-p2 I am using Ubuntu 10.04 DRBD - 8.3.11 In the first run I had everything working perfectly and when shifting the systems to a production environment I decided to restart the Meta Data creation part and start from scratch. The IPs had changed entirely in the production environment. Issuing drdbadm create-md r0 in both the servers runs successfully. But when I do "drbdadm -- --overwrite-data-of-peer primary all" on the primary it fails to start the re sync. My config file is as given below resource r0 { protocol C; syncer { rate 50M; } startup { wfc-timeout 15; degr-wfc-timeout 60; } net { cram-hmac-alg sha1; shared-secret "aklsadkjlhdbskjndsf8738734jkfkjfkjf"; } on primaryds { device /dev/drbd0; disk /dev/md2; address 172.16.7.1:7788; meta-disk internal; } on secondaryds { device /dev/drbd0; disk /dev/md2; address 172.16.7.3:7788; meta-disk internal; } } Status on primary root at primaryds:~# cat /proc/drbd version: 8.3.7 (api:88/proto:86-91) GIT-hash: ea9e28dbff98e331a62bcbcc63a6135808fe2917 build by root at primaryds, 2012-05-12 15:08:01 0: cs:WFBitMapS ro:Primary/Secondary ds:UpToDate/Inconsistent C r---- ns:0 nr:0 dw:0 dr:200 al:0 bm:0 lo:0 pe:0 ua:0 ap:0 ep:1 wo:b oos:5690352828 Status on secondary root at secondaryds:/etc/drbd.d# cat /proc/drbd version: 8.3.7 (api:88/proto:86-91) GIT-hash: ea9e28dbff98e331a62bcbcc63a6135808fe2917 build by root at secondaryds, 2012-05-12 15:25:25 0: cs:WFBitMapT ro:Secondary/Primary ds:Inconsistent/UpToDate C r---- ns:0 nr:0 dw:0 dr:0 al:0 bm:0 lo:0 pe:0 ua:0 ap:0 ep:1 wo:b oos:5690352828 Log of Primary May 30 13:42:23 primaryds kernel: [ 1584.057076] block drbd0: role( Secondary -> Primary ) disk( Inconsistent -> UpToDate ) May 30 13:42:23 primaryds kernel: [ 1584.086264] block drbd0: Forced to consider local data as UpToDate! May 30 13:42:23 primaryds kernel: [ 1584.086303] block drbd0: Creating new current UUID May 30 13:42:26 primaryds kernel: [ 1586.405551] block drbd0: drbd_sync_handshake: May 30 13:42:26 primaryds kernel: [ 1586.405564] block drbd0: self E8A075F378173D4B:0000000000000004:0000000000000000:0000000000000000 bits:1422588207 flags:0 May 30 13:42:26 primaryds kernel: [ 1586.405574] block drbd0: peer 0000000000000004:0000000000000000:0000000000000000:0000000000000000 bits:1422588207 flags:0 May 30 13:42:26 primaryds kernel: [ 1586.405582] block drbd0: uuid_compare()=2 by rule 30 May 30 13:42:26 primaryds kernel: [ 1586.405587] block drbd0: Becoming sync source due to disk states. May 30 13:42:26 primaryds kernel: [ 1586.405592] block drbd0: Writing the whole bitmap, full sync required after drbd_sync_handshake. May 30 13:42:27 primaryds kernel: [ 1588.171638] block drbd0: 5427 GB (1422588207 bits) marked out-of-sync by on disk bit-map. May 30 13:42:27 primaryds kernel: [ 1588.172769] block drbd0: conn( Connected -> WFBitMapS ) Log in Secondary May 30 13:42:24 secondaryds kernel: [ 1563.304894] block drbd0: peer( Secondary - Primary ) pdsk( Inconsistent - UpToDate ) May 30 13:42:24 secondaryds kernel: [ 1563.339674] block drbd0: drbd_sync_handshake: May 30 13:42:24 secondaryds kernel: [ 1563.339685] block drbd0: self 0000000000000004:0000000000000000:0000000000000000:0000000000000000 bits:1422588207 flags:0 May 30 13:42:24 secondaryds kernel: [ 1563.339695] block drbd0: peer E8A075F378173D4B:0000000000000004:0000000000000000:0000000000000000 bits:1422588207 flags:0 May 30 13:42:24 secondaryds kernel: [ 1563.339703] block drbd0: uuid_compare()=-2 by rule 20 May 30 13:42:24 secondaryds kernel: [ 1563.339709] block drbd0: Becoming sync target due to disk states. May 30 13:42:24 secondaryds kernel: [ 1563.339714] block drbd0: Writing the whole bitmap, full sync required after drbd_sync_handshake. May 30 13:42:26 secondaryds kernel: [ 1565.652342] block drbd0: 5427 GB (1422588207 bits) marked out-of-sync by on disk bit-map. May 30 13:42:26 secondaryds kernel: [ 1565.652965] block drbd0: conn( Connected - WFBitMapT ) The serves are not responding once it reaches this stage. Tried redoing it couple of time but noting happens. Why could the resync not be taking place? I would like some advice? Directions?

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  • SQL Server 2012 : The Data Tools installer is now available

    - by AaronBertrand
    Last week when RC0 was released, the updated installer for "Juneau" (SQL Server Data Tools) was not available. Depending on how you tried to get it, you either ended up on a blank search page, or a page offering the CTP3 bits. Important note: the CTP3 Juneau bits are not compatible with SQL Server 2012 RC0. If you already have Visual Studio 2010 installed (meaning Standard/Pro/Premium/Ultimate), you will need to install Service Pack 1 before continuing. You can get to the installer simply by opening...(read more)

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  • Ralink rt3090 driver installed and wireless doesn't work on Ubuntu 10.04

    - by Marcus Rene
    I have a LG A-410 lap-top (64 bits) with rt 3090 wireless card. Searching the problem I discover that I already have a rt 3090-dkms installed, but my wireless doesn't work. *-network UNCLAIMED description: Network controller product: RT3090 Wireless 802.11n 1T/1R PCIe vendor: RaLink physical id: 0 bus info: pci@0000:02:00.0 version: 00 width: 32 bits clock: 33MHz capabilities: pm msi pciexpress bus_master cap_list configuration: latency=0 resources: memory:e5400000-e540ffff

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  • Cannot get Virtualbox to install properly on Ubuntu 12.04

    - by lopac1029
    I cannot get Virtualbox to install properly on my 12.04. I first went with a manual install for the .deb from the old builds section of the Virtualbox page. That .deb opened up the Software Center and installed. Then I got the error coming up of VT-x/AMD-V hardware acceleration is not available on your system. Your 64-bit guest will fail to detect a 64-bit CPU and will not be able to boot. which I can only assume was due to my Ubuntu version being 32-bit (System Details - Overview - OC type: 32-bit, right?) So I followed these instructions to remove the .deb manually, restarted my laptop, and then FOUND the actual Virtualbox install in the Software Center and installed from that (assuming it would give me the correct version I need for my system) So after all that (and then some), I'm still getting the same error when I connect to my new job's project in Virtualbox. Can anyone point me in the right direction of what to do here? This is the first time I've ever worked with Virtualbox, and no one at this company is using Ubuntu, so I'm on my own here. EDIT: Here is the direct info from running the 2 suggested commands Inspiron-1750-brick:~ $lscpu Architecture: i686 CPU op-mode(s): 32-bit, 64-bit Byte Order: Little Endian CPU(s): 2 On-line CPU(s) list: 0,1 Thread(s) per core: 1 Core(s) per socket: 2 Socket(s): 1 Vendor ID: GenuineIntel CPU family: 6 Model: 23 Stepping: 10 CPU MHz: 2100.000 BogoMIPS: 4189.45 L1d cache: 32K L1i cache: 32K L2 cache: 2048K Inspiron-1750-brick:~ $cat /proc/cpuinfo processor : 0 vendor_id : GenuineIntel cpu family : 6 model : 23 model name : Intel(R) Core(TM)2 Duo CPU T6500 @ 2.10GHz stepping : 10 microcode : 0xa07 cpu MHz : 1200.000 cache size : 2048 KB physical id : 0 siblings : 2 core id : 0 cpu cores : 2 apicid : 0 initial apicid : 0 fdiv_bug : no hlt_bug : no f00f_bug : no coma_bug : no fpu : yes fpu_exception : yes cpuid level : 13 wp : yes flags : fpu vme de pse tsc msr pae mce cx8 apic sep mtrr pge mca cmov pat pse36 clflush dts acpi mmx fxsr sse sse2 ss ht tm pbe nx lm constant_tsc arch_perfmon pebs bts aperfmperf pni dtes64 monitor ds_cpl est tm2 ssse3 cx16 xtpr pdcm sse4_1 xsave lahf_lm dtherm bogomips : 4189.80 clflush size : 64 cache_alignment : 64 address sizes : 36 bits physical, 48 bits virtual power management: processor : 1 vendor_id : GenuineIntel cpu family : 6 model : 23 model name : Intel(R) Core(TM)2 Duo CPU T6500 @ 2.10GHz stepping : 10 microcode : 0xa07 cpu MHz : 1200.000 cache size : 2048 KB physical id : 0 siblings : 2 core id : 1 cpu cores : 2 apicid : 1 initial apicid : 1 fdiv_bug : no hlt_bug : no f00f_bug : no coma_bug : no fpu : yes fpu_exception : yes cpuid level : 13 wp : yes flags : fpu vme de pse tsc msr pae mce cx8 apic sep mtrr pge mca cmov pat pse36 clflush dts acpi mmx fxsr sse sse2 ss ht tm pbe nx lm constant_tsc arch_perfmon pebs bts aperfmperf pni dtes64 monitor ds_cpl est tm2 ssse3 cx16 xtpr pdcm sse4_1 xsave lahf_lm dtherm bogomips : 4189.45 clflush size : 64 cache_alignment : 64 address sizes : 36 bits physical, 48 bits virtual power management:

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  • How to know the number of pins on my laptop's ram?

    - by Rajat Saxena
    I am thinking about upgrading my laptop's ram.How can I get to know the number of pins on my ram without opening my laptop? I ran this command sudo dmidecode --type memory and got following info: Handle 0x0019, DMI type 17, 27 bytes Memory Device Array Handle: 0x0017 Error Information Handle: 0x001A Total Width: 64 bits Data Width: 64 bits Size: 1024 MB Form Factor: SODIMM Set: None Locator: DIMM0 Bank Locator: BANK 0 Type: DDR2 Type Detail: Synchronous Speed: 667 MHz Manufacturer: AD00000000000000 Serial Number: 04008104 Asset Tag: Unknown Part Number: 040404040404040404040404040404040404 Can anyone help?

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  • Why does cpuinfo report that my frequency is slower?

    - by Avery Chan
    My machine is running off of a AMD Sempron(tm) X2 190 Processor. According the marketing copy, it should be running at around 2.5 Ghz. Why is the cpu speed being reported as something lower? Spec description (in Chinese) $ cat /proc/cpuinfo processor : 0 vendor_id : AuthenticAMD cpu family : 16 model : 6 model name : AMD Sempron(tm) X2 190 Processor stepping : 3 microcode : 0x10000c8 cpu MHz : 800.000 cache size : 512 KB physical id : 0 siblings : 2 core id : 0 cpu cores : 2 apicid : 0 initial apicid : 0 fpu : yes fpu_exception : yes cpuid level : 5 wp : yes flags : fpu vme de pse tsc msr pae mce cx8 apic sep mtrr pge mca cmov pat pse36 clflush mmx fxsr sse sse2 ht syscall nx mmxext fxsr_opt pdpe1gb rdtscp lm 3dnowext 3dnow constant_tsc rep_good nopl nonstop_tsc extd_apicid pni monitor cx16 popcnt lahf_lm cmp_legacy svm extapic cr8_legacy abm sse4a misalignsse 3dnowprefetch osvw ibs skinit wdt npt lbrv svm_lock nrip_save bogomips : 5022.89 TLB size : 1024 4K pages clflush size : 64 cache_alignment : 64 address sizes : 48 bits physical, 48 bits virtual power management: ts ttp tm stc 100mhzsteps hwpstate processor : 1 vendor_id : AuthenticAMD cpu family : 16 model : 6 model name : AMD Sempron(tm) X2 190 Processor stepping : 3 microcode : 0x10000c8 cpu MHz : 800.000 cache size : 512 KB physical id : 0 siblings : 2 core id : 1 cpu cores : 2 apicid : 1 initial apicid : 1 fpu : yes fpu_exception : yes cpuid level : 5 wp : yes flags : fpu vme de pse tsc msr pae mce cx8 apic sep mtrr pge mca cmov pat pse36 clflush mmx fxsr sse sse2 ht syscall nx mmxext fxsr_opt pdpe1gb rdtscp lm 3dnowext 3dnow constant_tsc rep_good nopl nonstop_tsc extd_apicid pni monitor cx16 popcnt lahf_lm cmp_legacy svm extapic cr8_legacy abm sse4a misalignsse 3dnowprefetch osvw ibs skinit wdt npt lbrv svm_lock nrip_save bogomips : 5022.82 TLB size : 1024 4K pages clflush size : 64 cache_alignment : 64 address sizes : 48 bits physical, 48 bits virtual power management: ts ttp tm stc 100mhzsteps hwpstate

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