bit-manipulation

问题 I'm interested in a fast method for "expanding bits," which can be defined as the following: Let B be a binary number with n bits, i.e. B \in {0,1}^ n Let P be the position of all 1/true bits in B , i.e. 1 << p[i] & B == 1 , and | P |= k For another given number, A \in {0,1}^ k , let Ap be the bit-expanded form of A given B , such that Ap[j] == A[j] << p[j] . The result of the "bit expansion" is Ap . A couple examples: Given B : 00 1 0 111 0, A : 0110, then Ap should be 00 0 0 110 0 Given B :

问题 I am trying to replicate the behavior of javascript bit-shift and bit-wise operations in java. Did you ever try to do this before, and how can you do it reliably and consistently even with longs? var i=[some array with large integers]; for(var x=0;x<100;x++) { var a=a large integer; var z=some 'long'>2.1 billion; //EDIT: z=i[x]+=(z>>>5)^(a<<2))+((z<<4)^(a<<5)); } what would you do to put this into java? 回答1: Yes. Java has bit-wise operators and shift operators. Is there something in

问题 I am trying to eliminate an IF statement whereby if I receive the number 32 I would like a '1', but any other number I would like a '0'. 32 is 0010 0000 so I thought about XOR-ing my input number with 1101 1111. Therefore if I get the number 32 I end up with 1111 1111. Now is there any way of AND-ing the individual bits (1111 1111), because if one of my XOR results is a 0, it means my final AND-ed value is 0, otherwise its a 1? EDIT: Using GCC, not Intel compiler (because I know there are a

问题 This question already has answers here : what does it mean to bitwise left shift an unsigned char with 16 (2 answers) Closed 12 months ago . I am trying to understand shift operators in C/C++, but they are giving me a tough time. I have an unsigned 8-bit integer initialized to a value, for the example, say 1. uint8_t x = 1; From my understanding, it is represented in the memory like |0|0|0|0|0||0||0||1| . Now, when I am trying to left shit the variable x by 16 bit, I am hoping to get output 0

问题 This question already has answers here : Bit twiddling: which bit is set? (15 answers) Closed 7 years ago . I am trying to find an optimal code to locate a single bit index in a long integer (64 bit). The long integer has exactly one set bit. (Using C language) Currently, I am just shifting the whole thing by one bit, then checking for zero. I have read about lookup tables, but it won't do for the whole 64 bits. I thought about checking each 8 bits for zero, if not use a lookup, but still I

问题 I've come across some code which has the bit masks 0xff and 0xff00 or in 16 bit binary form 00000000 11111111 and 11111111 00000000 . /** * Function to check if the given string is in GZIP Format. * * @param inString String to check. * @return True if GZIP Compressed otherwise false. */ public static boolean isStringCompressed(String inString) { try { byte[] bytes = inString.getBytes("ISO-8859-1"); int gzipHeader = ((int) bytes[0] & 0xff) | ((bytes[1] << 8) & 0xff00); return GZIPInputStream

问题 I'm trying to blend two colors that are coded as integers. Here is my little function: int blend (int a, int b, float ratio) { if (ratio > 1f) { ratio = 1f; } else if (ratio < 0f) { ratio = 0f; } float iRatio = 1.0f - ratio; int aA = (a >> 24 & 0xff); int aR = ((a & 0xff0000) >> 16); int aG = ((a & 0xff00) >> 8); int aB = (a & 0xff); int bA = (b >> 24 & 0xff); int bR = ((b & 0xff0000) >> 16); int bG = ((b & 0xff00) >> 8); int bB = (b & 0xff); int A = ((int)(aA * iRatio) + (int)(bA * ratio));

问题 I'm trying to blend two colors that are coded as integers. Here is my little function: int blend (int a, int b, float ratio) { if (ratio > 1f) { ratio = 1f; } else if (ratio < 0f) { ratio = 0f; } float iRatio = 1.0f - ratio; int aA = (a >> 24 & 0xff); int aR = ((a & 0xff0000) >> 16); int aG = ((a & 0xff00) >> 8); int aB = (a & 0xff); int bA = (b >> 24 & 0xff); int bR = ((b & 0xff0000) >> 16); int bG = ((b & 0xff00) >> 8); int bB = (b & 0xff); int A = ((int)(aA * iRatio) + (int)(bA * ratio));

问题 Many lossless algorithms in signal processing require an evaluation of the expression of the form ⌊ a / 2 b ⌋, where a , b are signed ( a possibly negative, b non-negative) integers and ⌊·⌋ is the floor function. This usually leads to the following implementation. int floor_div_pow2(int numerator, int log2_denominator) { return numerator >> log2_denominator; } Unfortunately, the C standard states that the result of the >> operator is implementation-defined if the left operand has a signed

问题 I am not sure of the precise definition of this term. I know that a bitwise XOR operation is going bit by bit and taking the XOR of corresponding bits position wise. Is this result termed the 'XOR sum'? If not, what is an XOR sum, and how do you use XOR to implement this addition? 回答1: In a bit wise XOR operation: a b a^b ----------- 0 0 0 0 1 1 1 0 1 1 1 0 XOR sum refers to successive XOR operations on integers. Suppose you have numbers from 1 to N and you have to find their XOR sum then for