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  • Detection of negative integers using bit operations

    - by Nawaz
    One approach to check if a given integer is negative or not, could be this: (using bit operations) int num_bits = sizeof(int) * 8; //assuming 8 bits per byte! int sign_bit = given_int & (1 << (num_bits-1)); //sign_bit is either 1 or 0 if ( sign_bit ) { cout << "given integer is negative"<<endl; } else { cout << "given integer is positive"<<endl; } The problem with this solution is that number of bits per byte couldn't be 8, it could be 9,10, 11 even 16 or 40 bits per byte. Byte doesn't necessarily mean 8 bits! Anyway, this problem can be easily fixed by writing, //CHAR_BIT is defined in limits.h int num_bits = sizeof(int) * CHAR_BIT; //no assumption. It seems fine now. But is it really? Is this Standard conformant? What if the negative integer is not represented as 2's complement? What if it's representation in a binary numeration system that doesn't necessitate only negative integers to have 1 in it's most significant bit? Can we write such code that will be both portable and standard conformant? Related topics: Size of Primitive data types Why is a boolean 1 byte and not 1 bit of size?

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  • Good way to identify similar images?

    - by Nick
    I've developed a simple and fast algorithm in PHP to compare images for similarity. Its fast (~40 per second for 800x600 images) to hash and a unoptimised search algorithm can go through 3,000 images in 22 mins comparing each one against the others (3/sec). The basic overview is you get a image, rescale it to 8x8 and then convert those pixels for HSV. The Hue, Saturation and Value are then truncated to 4 bits and it becomes one big hex string. Comparing images basically walks along two strings, and then adds the differences it finds. If the total number is below 64 then its the same image. Different images are usually around 600 - 800. Below 20 and extremely similar. Are there any improvements upon this model I can use? I havent looked at how relevant the different components (hue, saturation and value) are to the comparison. Hue is probably quite important but the others? To speed up searches I could probably split the 4 bits from each part in half, and put the most significant bits first so if they fail the check then the lsb doesnt need to be checked at all. I dont know a efficient way to store bits like that yet still allow them to be searched and compared easily. I've been using a dataset of 3,000 photos (mostly unique) and there havent been any false positives. Its completely immune to resizes and fairly resistant to brightness and contrast changes.

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  • python crc32 woes

    - by lazyr
    I'm writing a python program to extract data from the middle of a 6 GB bz2 file. A bzip2 file is made up of independently decryptable blocks of data, so I only need to find a block (they are delimited by magic bits), then create a temporary one-block bzip2 file from it in memory, and finally pass that to the bz2.decompress function. Easy, no? The bzip2 format has a crc32 checksum for the file at the end. No problem, binascii.crc32 to the rescue. But wait. The data to be checksummed does not necessarily end on a byte boundary, and the crc32 function operates on a whole number of bytes. My plan: use the binascii.crc32 function on all but the last byte, and then a function of my own to update the computed crc with the last 1-7 bits. But hours of coding and testing has left me bewildered, and my puzzlement can be boiled down to this question: how come crc32("\x00") is not 0x00000000? Shouldn't it be, according to the wikipedia article? You start with 0b00000000 and pad with 32 0's, then do polynomial division with 0x04C11DB7 until there are no ones left in the first 8 bits, which is immediately. Your last 32 bits is the checksum, and how can that not be all zeroes? I've searched google for answers and looked at the code of several crc32 implementations without finding any clue to why this is so.

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  • Change default java installation

    - by user1501700
    I have many Java versions installed on a Windows 7 machine. Some of them are 32 bits, some are 64 bits. Now as default it starts one of those last versions (1.7 64 bits). How do I tell my Windows 7 machine to use another version of Java? One of the reasons is that I'm developing a JNI project from Microsoft Visual Studio C++ - it uses also java 1.7 64 bits. Best regards, Andrej I have set: User variable: JAVA_HOME=C:\j2sdk1.4.2_04 PATH=%JAVA_HOME%\bin;%PATH% and system variable: JAVA_HOME=C:\j2sdk1.4.2_04 PATH=...a_lot_of_paths...;%JAVA_HOME%\bin;%PATH% I had no idea which is better to set - for user or system settings. Done both. System restart. And...it didn't helped :( When I run "java -version" from cmd i have java 1.7, but not java 1.4 like defined in PATH. after run C:where java I got two results: C:\Windows\System32\java.exe C:\j2sdk1.4.2_04\bin\java.exe Who let Java go to my windows directory ???!!! How to deal with that?

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  • What's up with this reversing bit order function?

    - by MattyW
    I'm rather ashamed to admit that I don't know as much about bits and bit manipulation as I probably should. I tried to fix that this weekend by writing some 'reverse the order of bits' and 'count the ON bits' functions. I took an example from here but when I implemented it as below, I found I had to be looping while < 29. If I loop while < 32 (as in the example) Then when I try to print the integer (using a printBits function i've written) I seem to be missing the first 3 bits. This makes no sense to me, can someone help me out? int reverse(int n) { int r = 0; int i = 0; for(i = 0; i < 29; i++) { r = (r << 1) + (n & 1); n >>=1; } return r; }

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  • How to install Archlinux on a acer emachines em350 ?

    - by zillion
    Do a 64 bits version will work better if the cpu is 64 bits capable ? Can I remove all added stuffs on Archbang or CTKArchLive to return to plain Archlinux with openbox that I can configure myself easily, so I can install Archlinux faster ? I tried and installed Arch properly but I need to install the broadcom bcm 4727 driver (b43,wl,sta,open source), which driver do I need and how can I install it by my usb on Archlinux CLI made with base, sudo and wireless_tools only ?

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  • Ubuntu 12.04 LTS vs Ubuntu 14.04 LTS memory usage

    - by geoffroy
    My droplet has 512 MB memory and is running Ubuntu 12.04 LTS 64 bits and a Rails 4 application + several workers. It's running well. I tried to deploy the same thing on a Ubuntu 14.04 LTS 64 bits droplet and I've got plenty of memory related problem (can't fork). Is Ubuntu 14.04 LTS using way more RAM than Ubuntu 12.04 LTS? Is there something I should know to lower memory usage ? Should I stick with Ubuntu 12.04 LTS?

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  • Newly configured MSSQL2008, TIME_WAIT but no ESTABLISHED?

    - by 3molo
    Windows 2008 R2, standard. No firewall locally on it. Newly setup because an old SQL2000 had two disks die (or could it be the raid controller?) at the same time. Luckily, I had fresh backups. The databases have been restored, and SP2 for SQL2008 applied. I can see various hosts trying to establish a session, but the (customer) sites does not work and I don't see the expected established sessions. A wireshark reveals a full three-way handshake. Since it's customer machines connecting, I cannot logon to them and restart application pools.. What on earth could be causing this? No. Time Source Destination Protocol Info 1 0.000000 1.2.5.127 1.2.6.133 TCP desktop-dna > ms-sql-s [SYN] Seq=0 Win=65535 Len=0 MSS=1380 SACK_PERM=1 Frame 1: 62 bytes on wire (496 bits), 62 bytes captured (496 bits) Ethernet II, Src: Cisco_31:5e:09 (00:26:0b:31:5e:09), Dst: Vmware_b7:00:05 (00:50:56:b7:00:05) Internet Protocol, Src: 1.2.5.127 (1.2.5.127), Dst: 1.2.6.133 (1.2.6.133) Transmission Control Protocol, Src Port: desktop-dna (2763), Dst Port: ms-sql-s (1433), Seq: 0, Len: 0 No. Time Source Destination Protocol Info 2 0.000123 1.2.6.133 1.2.5.127 TCP ms-sql-s > desktop-dna [SYN, ACK] Seq=0 Ack=1 Win=8192 Len=0 MSS=1460 SACK_PERM=1 Frame 2: 62 bytes on wire (496 bits), 62 bytes captured (496 bits) Ethernet II, Src: Vmware_b7:00:05 (00:50:56:b7:00:05), Dst: Cisco_31:5e:09 (00:26:0b:31:5e:09) Internet Protocol, Src: 1.2.6.133 (1.2.6.133), Dst: 1.2.5.127 (1.2.5.127) Transmission Control Protocol, Src Port: ms-sql-s (1433), Dst Port: desktop-dna (2763), Seq: 0, Ack: 1, Len: 0 No. Time Source Destination Protocol Info 3 0.000884 1.2.5.127 1.2.6.133 TCP desktop-dna > ms-sql-s [ACK] Seq=1 Ack=1 Win=65535 Len=0 And netstat TCP 1.2.6.133:1433 1.2.2.98:26895 TIME_WAIT 0 TCP 1.2.6.133:1433 1.2.2.98:26912 TIME_WAIT 0 TCP 1.2.6.133:1433 1.2.2.98:26918 TIME_WAIT 0 TCP 1.2.6.133:1433 1.2.2.98:26931 TIME_WAIT 0 TCP 1.2.6.133:1433 1.2.5.127:2736 TIME_WAIT 0 TCP 1.2.6.133:1433 1.2.5.127:2737 TIME_WAIT 0 TCP 1.2.6.133:1433 1.2.5.127:2738 TIME_WAIT 0 TCP 1.2.6.133:1433 1.2.5.127:2739 TIME_WAIT 0

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  • Per bytes RAM memory acess

    - by b-gen-jack-o-neill
    Hi, I have just a simple question. Today memory DDR chips are 64 bits wide, and the CPU data bus is also 64 bits wide. But memory is stil organised in single bytes. So, what I want to ask is, when CPU selects some memory adress, it should be one byte, right? Becouse the lowest memory portion you can access is 1 byte. But, if you get 1 byte per 1 adress, why is memory bus 8 bytes wide?

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  • ubuntu 64 or 32 bit for macbook/vps?

    - by ajsie
    i've got macbook pro and wonder if i should use 64 or 32 bits ubuntu server? and then i've got a vps not hosted by med. how do i know what version to choose? how do you check how many bits your cpu i working with? can i use 64 on 32 and vice versa?

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  • Can you help me get my head around openssl public key encryption with rsa.h in c++?

    - by Ben
    Hi there, I am trying to get my head around public key encryption using the openssl implementation of rsa in C++. Can you help? So far these are my thoughts (please do correct if necessary) Alice is connected to Bob over a network Alice and Bob want secure communications Alice generates a public / private key pair and sends public key to Bob Bob receives public key and encrypts a randomly generated symmetric cypher key (e.g. blowfish) with the public key and sends the result to Alice Alice decrypts the ciphertext with the originally generated private key and obtains the symmetric blowfish key Alice and Bob now both have knowledge of symmetric blowfish key and can establish a secure communication channel Now, I have looked at the openssl/rsa.h rsa implementation (since I already have practical experience with openssl/blowfish.h), and I see these two functions: int RSA_public_encrypt(int flen, unsigned char *from, unsigned char *to, RSA *rsa, int padding); int RSA_private_decrypt(int flen, unsigned char *from, unsigned char *to, RSA *rsa, int padding); If Alice is to generate *rsa, how does this yield the rsa key pair? Is there something like rsa_public and rsa_private which are derived from rsa? Does *rsa contain both public and private key and the above function automatically strips out the necessary key depending on whether it requires the public or private part? Should two unique *rsa pointers be generated so that actually, we have the following: int RSA_public_encrypt(int flen, unsigned char *from, unsigned char *to, RSA *rsa_public, int padding); int RSA_private_decrypt(int flen, unsigned char *from, unsigned char *to, RSA *rsa_private, int padding); Secondly, in what format should the *rsa public key be sent to Bob? Must it be reinterpreted in to a character array and then sent the standard way? I've heard something about certificates -- are they anything to do with it? Sorry for all the questions, Best Wishes, Ben. EDIT: Coe I am currently employing: /* * theEncryptor.cpp * * * Created by ben on 14/01/2010. * Copyright 2010 __MyCompanyName__. All rights reserved. * */ #include "theEncryptor.h" #include <iostream> #include <sys/socket.h> #include <sstream> theEncryptor::theEncryptor() { } void theEncryptor::blowfish(unsigned char *data, int data_len, unsigned char* key, int enc) { // hash the key first! unsigned char obuf[20]; bzero(obuf,20); SHA1((const unsigned char*)key, 64, obuf); BF_KEY bfkey; int keySize = 16;//strlen((char*)key); BF_set_key(&bfkey, keySize, obuf); unsigned char ivec[16]; memset(ivec, 0, 16); unsigned char* out=(unsigned char*) malloc(data_len); bzero(out,data_len); int num = 0; BF_cfb64_encrypt(data, out, data_len, &bfkey, ivec, &num, enc); //for(int i = 0;i<data_len;i++)data[i]=out[i]; memcpy(data, out, data_len); free(out); } void theEncryptor::generateRSAKeyPair(int bits) { rsa = RSA_generate_key(bits, 65537, NULL, NULL); } int theEncryptor::publicEncrypt(unsigned char* data, unsigned char* dataEncrypted,int dataLen) { return RSA_public_encrypt(dataLen, data, dataEncrypted, rsa, RSA_PKCS1_OAEP_PADDING); } int theEncryptor::privateDecrypt(unsigned char* dataEncrypted, unsigned char* dataDecrypted) { return RSA_private_decrypt(RSA_size(rsa), dataEncrypted, dataDecrypted, rsa, RSA_PKCS1_OAEP_PADDING); } void theEncryptor::receivePublicKeyAndSetRSA(int sock, int bits) { int max_hex_size = (bits / 4) + 1; char keybufA[max_hex_size]; bzero(keybufA,max_hex_size); char keybufB[max_hex_size]; bzero(keybufB,max_hex_size); int n = recv(sock,keybufA,max_hex_size,0); n = send(sock,"OK",2,0); n = recv(sock,keybufB,max_hex_size,0); n = send(sock,"OK",2,0); rsa = RSA_new(); BN_hex2bn(&rsa->n, keybufA); BN_hex2bn(&rsa->e, keybufB); } void theEncryptor::transmitPublicKey(int sock, int bits) { const int max_hex_size = (bits / 4) + 1; long size = max_hex_size; char keyBufferA[size]; char keyBufferB[size]; bzero(keyBufferA,size); bzero(keyBufferB,size); sprintf(keyBufferA,"%s\r\n",BN_bn2hex(rsa->n)); sprintf(keyBufferB,"%s\r\n",BN_bn2hex(rsa->e)); int n = send(sock,keyBufferA,size,0); char recBuf[2]; n = recv(sock,recBuf,2,0); n = send(sock,keyBufferB,size,0); n = recv(sock,recBuf,2,0); } void theEncryptor::generateRandomBlowfishKey(unsigned char* key, int bytes) { /* srand( (unsigned)time( NULL ) ); std::ostringstream stm; for(int i = 0;i<bytes;i++){ int randomValue = 65 + rand()% 26; stm << (char)((int)randomValue); } std::string str(stm.str()); const char* strs = str.c_str(); for(int i = 0;bytes;i++)key[i]=strs[i]; */ int n = RAND_bytes(key, bytes); if(n==0)std::cout<<"Warning key was generated with bad entropy. You should not consider communication to be secure"<<std::endl; } theEncryptor::~theEncryptor(){}

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  • Tracking unique versions of files with hashes

    - by rwmnau
    I'm going to be tracking different versions of potentially millions of different files, and my intent is to hash them to determine I've already seen that particular version of the file. Currently, I'm only using MD5 (the product is still in development, so it's never dealt with millions of files yet), which is clearly not long enough to avoid collisions. However, here's my question - Am I more likely to avoid collisions if I hash the file using two different methods and store both hashes (say, SHA1 and MD5), or if I pick a single, longer hash (like SHA256) and rely on that alone? I know option 1 has 288 hash bits and option 2 has only 256, but assume my two choices are the same total hash length. Since I'm dealing with potentially millions of files (and multiple versions of those files over time), I'd like to do what I can to avoid collisions. However, CPU time isn't (completely) free, so I'm interested in how the community feels about the tradeoff - is adding more bits to my hash proportionally more expensive to compute, and are there any advantages to multiple different hashes as opposed to a single, longer hash, given an equal number of bits in both solutions?

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  • Java bitshift strangeness

    - by Martin
    Java has 2 bitshift operators for right shifts: >> shifts right, and is dependant on the sign bit for the sign of the result >>> shifts right and shifts a zero into leftmost bits http://java.sun.com/docs/books/tutorial/java/nutsandbolts/op3.html This seems fairly simple, so can anyone explain to me why this code, when given a value of -128 for bar, produces a value of -2 for foo: byte foo = (byte)((bar & ((byte)-64)) >>> 6); What this is meant to do is take an 8bit byte, mask of the leftmost 2 bits, and shift them into the rightmost 2 bits. Ie: initial = 0b10000000 (-128) -64 = 0b11000000 initial & -64 = 0b10000000 0b10000000 >>> 6 = 0b00000010 The result actually is -2, which is 0b11111110 Ie. 1s rather than zeros are shifted into left positions

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  • problems in trying ieee 802.15.4 working from msk

    - by asel
    Hi, i took a msk code from dsplog.com and tried to modify it to test the ieee 802.15.4. There are several links on that site for ieee 802.15.4. Currently I am getting simulated ber results all approximately same for all the cases of Eb_No values. Can you help me to find why? thanks in advance! clear PN = [ 1 1 0 1 1 0 0 1 1 1 0 0 0 0 1 1 0 1 0 1 0 0 1 0 0 0 1 0 1 1 1 0; 1 1 1 0 1 1 0 1 1 0 0 1 1 1 0 0 0 0 1 1 0 1 0 1 0 0 1 0 0 0 1 0; 0 0 1 0 1 1 1 0 1 1 0 1 1 0 0 1 1 1 0 0 0 0 1 1 0 1 0 1 0 0 1 0; 0 0 1 0 0 0 1 0 1 1 1 0 1 1 0 1 1 0 0 1 1 1 0 0 0 0 1 1 0 1 0 1; 0 1 0 1 0 0 1 0 0 0 1 0 1 1 1 0 1 1 0 1 1 0 0 1 1 1 0 0 0 0 1 1; 0 0 1 1 0 1 0 1 0 0 1 0 0 0 1 0 1 1 1 0 1 1 0 1 1 0 0 1 1 1 0 0; 1 1 0 0 0 0 1 1 0 1 0 1 0 0 1 0 0 0 1 0 1 1 1 0 1 1 0 1 1 0 0 1; 1 0 0 1 1 1 0 0 0 0 1 1 0 1 0 1 0 0 1 0 0 0 1 0 1 1 1 0 1 1 0 1; 1 0 0 0 1 1 0 0 1 0 0 1 0 1 1 0 0 0 0 0 0 1 1 1 0 1 1 1 1 0 1 1; 1 0 1 1 1 0 0 0 1 1 0 0 1 0 0 1 0 1 1 0 0 0 0 0 0 1 1 1 0 1 1 1; 0 1 1 1 1 0 1 1 1 0 0 0 1 1 0 0 1 0 0 1 0 1 1 0 0 0 0 0 0 1 1 1; 0 1 1 1 0 1 1 1 1 0 1 1 1 0 0 0 1 1 0 0 1 0 0 1 0 1 1 0 0 0 0 0; 0 0 0 0 0 1 1 1 0 1 1 1 1 0 1 1 1 0 0 0 1 1 0 0 1 0 0 1 0 1 1 0; 0 1 1 0 0 0 0 0 0 1 1 1 0 1 1 1 1 0 1 1 1 0 0 0 1 1 0 0 1 0 0 1; 1 0 0 1 0 1 1 0 0 0 0 0 0 1 1 1 0 1 1 1 1 0 1 1 1 0 0 0 1 1 0 0; 1 1 0 0 1 0 0 1 0 1 1 0 0 0 0 0 0 1 1 1 0 1 1 1 1 0 1 1 1 0 0 0; ]; N = 5*10^5; % number of bits or symbols fsHz = 1; % sampling period T = 4; % symbol duration Eb_N0_dB = [0:10]; % multiple Eb/N0 values ct = cos(pi*[-T:N*T-1]/(2*T)); st = sin(pi*[-T:N*T-1]/(2*T)); for ii = 1:length(Eb_N0_dB) tx = []; % MSK Transmitter ipBit = round(rand(1,N/32)*15); for k=1:length(ipBit) sym = ipBit(k); tx = [tx PN((sym+1),1:end)]; end ipMod = 2*tx - 1; % BPSK modulation 0 -> -1, 1 -> 1 ai = kron(ipMod(1:2:end),ones(1,2*T)); % even bits aq = kron(ipMod(2:2:end),ones(1,2*T)); % odd bits ai = [ai zeros(1,T) ]; % padding with zero to make the matrix dimension match aq = [zeros(1,T) aq ]; % adding delay of T for Q-arm % MSK transmit waveform xt = 1/sqrt(T)*[ai.*ct + j*aq.*st]; % Additive White Gaussian Noise nt = 1/sqrt(2)*[randn(1,N*T+T) + j*randn(1,N*T+T)]; % white gaussian noise, 0dB variance % Noise addition yt = xt + 10^(-Eb_N0_dB(ii)/20)*nt; % additive white gaussian noise % MSK receiver % multiplying with cosine and sine waveforms xE = conv(real(yt).*ct,ones(1,2*T)); xO = conv(imag(yt).*st,ones(1,2*T)); bHat = zeros(1,N); bHat(1:2:end) = xE(2*T+1:2*T:end-2*T); % even bits bHat(2:2:end) = xO(3*T+1:2*T:end-T); % odd bits result=zeros(16,1); chiplen=32; seqstart=1; recovered = []; while(seqstart<length(bHat)) A = bHat(seqstart:seqstart+(chiplen-1)); for j=1:16 B = PN(j,1:end); result(j)=sum(A.*B); end [value,index] = max(result); recovered = [recovered (index-1)]; seqstart = seqstart+chiplen; end; %# create binary string - the 4 forces at least 4 bits bstr1 = dec2bin(ipBit,4); bstr2 = dec2bin(recovered,4); %# convert back to numbers (reshape so that zeros are preserved) out1 = str2num(reshape(bstr1',[],1))'; out2 = str2num(reshape(bstr2',[],1))'; % counting the errors nErr(ii) = size(find([out1 - out2]),2); end nErr/(length(ipBit)*4) % simulated ber theoryBer = 0.5*erfc(sqrt(10.^(Eb_N0_dB/10))) % theoretical ber

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  • Efficiently storing a list of prime numbers

    - by eSKay
    This article says: Every prime number can be expressed as 30k±1, 30k±7, 30k±11, or 30k±13 for some k. That means we can use eight bits per thirty numbers to store all the primes; a million primes can be compressed to 33,334 bytes "That means we can use eight bits per thirty numbers to store all the primes" This "eight bits per thirty numbers" would be for k, correct? But each k value will not necessarily take-up just one bit. Shouldn't it be eight k values instead? "a million primes can be compressed to 33,334 bytes" I am not sure how this is true. We need to indicate two things: VALUE of k (can be arbitrarily large) STATE from one of the eight states (-13,-11,-7,-1,1,7,11,13) I am not following how 33,334 bytes was arrived at, but I can say one thing: as the prime numbers become larger and larger in value, we will need more space to store the value of k. How, then can we fix it at 33,334 bytes?

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  • How do I compute the approximate entropy of a bit string?

    - by dreeves
    Is there a standard way to do this? Googling -- "approximate entropy" bits -- uncovers multiple academic papers but I'd like to just find a chunk of pseudocode defining the approximate entropy for a given bit string of arbitrary length. (In case this is easier said than done and it depends on the application, my application involves 16,320 bits of encrypted data (cyphertext). But encrypted as a puzzle and not meant to be impossible to crack. I thought I'd first check the entropy but couldn't easily find a good definition of such. So it seemed like a question that ought to be on StackOverflow! Ideas for where to begin with de-cyphering 16k random-seeming bits are also welcome...) See also this related question: http://stackoverflow.com/questions/510412/what-is-the-computer-science-definition-of-entropy

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  • What is the difference between a 32-bit and 64-bit processor?

    - by JJG
    I have been trying to read up on 32-bit and 64-bit processors (http://en.wikipedia.org/wiki/32-bit_processing). My understanding is that a 32-bit processor (like x86) has registers 32-bits wide. I'm not sure what that means. So it has special "memory spaces" that can store integer values up to 2^32? I don't want to sound stupid, but I have no idea about processors. I'm assuming 64-bits is, in general, better than 32-bits. Although my computer now (one year old, Win 7, Intel Atom) has a 32-bit processor.

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  • bitshift large strings for encoding QR Codes

    - by icekreaman
    As an example, suppose a QR Code data stream contains 55 data words (each one byte in length) and 15 error correction words (again one byte). The data stream begins with a 12 bit header and ends with four 0 bits. So, 12 + 4 bits of header/footer and 15 bytes of error correction, leaves me 53 bytes to hold 53 alphanumeric characters. The 53 bytes of data and 15 bytes of ec are supplied in a string of length 68 (str68). The problem seems simple enough - concatenate 2 bytes of (right-shifted) header data with str68 and then left shift the entire 70 bytes by 4 bits. This is the first time in many years of programming that I have ever needed to do something like this, I am a c and bit shifting noob, so please be gentle... I have done a little investigation and so far have not been able to figure out how to bitshift 70 bytes of data; any help would be greatly appreciated. Larger QR codes can hold 2000 bytes of data...

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  • Does Django tests run slower on the mac compared to linux?

    - by Thierry Lam
    I'm currently developing my Django projects on both: Mac OS X 10.5, 32 bit Ubuntu Server 9.10 64 bits (1 CPU, 512MB RAM) Both of the above OS are using: Python 2.6.4 Django 1.1.1 MySQL 5.1 Running 12 tests for one of my application take: Mac: 57.513s Linux: 30.935s EDIT: Mac Hardware Spec: MacBook Pro 2.2 GHz Intel Core 2 Duo 3GB RAM I'm running the Ubuntu OS on the same mac above through VMware Fusion 2.0.6. You might argue that Ubuntu Server 64 bits is faster but I have observed a similar speed difference on Ubuntu 8.10 32 bits desktop edition. Even if I turn off my linux VM and other mac applications, I still experience the slowness. Has anyone else experienced this Django test speed difference across those two OS?

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  • C++ : size of int, long, etc...

    - by Jérôme
    I'm looking for detailed informations regarding the size of basic C++ types. I know that it depends on the architecture (16 bits, 32 bits, 64 bits) and the compiler. But are there any standards ? I'm using Visual Studio 2008 on a 32 bit achitecture. Here is what I get : char : 1 byte short : 2 bytes int : 4 bytes long : 4 bytes float : 4 bytes double : 8 bytes I tried to find, without much success, reliable informations telling the sizes of char, short, int , long, double, float (and other types I don't think of) under different architecture and compiler.

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  • how many color combinations in a 24 bit image

    - by numerical25
    I am reading a book and I am not sure if its a mistake or I am misunderstanding the quote. It reads... Nowadays every PC you can buy has hardware that can render images with at least 16.7 million individual colors. Rather than have an array with thousands of color entries, the images instead contain explicit color values for each pixel. A 24-bit display, of course, uses 24 bits, or 3 bytes per pixel, for color information. This gives 1 byte, or 256 distinct values each, for red, green, and blue. This is generally called true color, because 256^3 (16.7 million) He says 1 byte is equal to 256 distinct values. 1 byte = 8 bits. 8^2 bits = 64 distinct colors right ?? It's not adding up right to me. I know it might be something simple to understand, but I don't understand.

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  • how many color combinations in a 24 bit image

    - by numerical25
    I am reading a book and I am not sure if its a mistake or I am misunderstanding the quote. It reads... Nowadays every PC you can buy has hardware that can render images with at least 16.7 million individual colors. Rather than have an array with thousands of color entries, the images instead contain explicit color values for each pixel. A 24-bit display, of course, uses 24 bits, or 3 bytes per pixel, for color information. This gives 1 byte, or 256 distinct values each, for red, green, and blue. This is generally called true color, because 256^3 (16.7 million) He says 1 byte is equal to 256 distinct values. 1 byte = 8 bits. 8^2 bits = 64 combinations of colors right ?? It's not adding up right to me. I know it might be something simple to understand, but I don't understand.

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  • Bitfield mask/operations with optional items

    - by user1560249
    I'm trying to find a way to handle several bitfield cases that include optional, required, and not allowed positions. yy?nnn?y 11000001 ?yyy?nnn 01110000 nn?yyy?n 00011100 ?nnn?yyy 00000111 In these four cases, the ? indicates that the bit can be either 1 or 0 while y indicates a 1 is required and n indicates that a 0 is required. The bits to the left/right of the required bits can be anything and the remaining bits must be 0. Is there a masking method I can use to test if an input bit set satisfies one of these cases?

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  • Null Pointer Exception with an array of bitsets

    - by p0ny
    could someone explain to me why the following results in a Null pointer Exception? And how to set a value for bitarray[0]? BitSet[] bitarray; bitarray= new BitSet[10]; bitarray[0].set(1); Also, why something like this work and not result in a pointer exception? BitSet[] bitarray = new BitSet[10]; BitSet bits = new BitSet(32); bits.set(1); bitarray[0] = bits; Thanks

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