bits

Im trying to change from big endian to little endian on a double. One way to go is to use double val, tmp = 5.55; ((unsigned int *)&val)[0] = ntohl(((unsigned int *)&tmp)[1]); ((unsigned int *)&val)[1] = ntohl(((unsigned int *)&tmp)[0]); But then I get a warning: "dereferencing type-punned pointer will break strict-aliasing rules" and I dont want to turn this warning off. Another way to go is: #define ntohll(x) ( ( (uint64_t)(ntohl( (uint32_t)((x << 32) >> 32) )) << 32) | ntohl( ((uint32_t)(x >> 32)) ) ) val = (double)bswap_64(unsigned long long(tmp)); //or val = (double)ntohll(unsigned long

What is wrong with the below code which prints the binary representation of a number? int a = 65; for (int i = 0; i < 8; i++) { cout << ((a >> i) & 1); } You're starting at the least significant bit in the number and printing it first. However, whatever you print first is the most significant digit in the typical binary representation. 65 is 01000001 so this is how your loop iterates 01000001 ^ Output: 1 01000001 ^ Output: 10 01000001 ^ Output: 100 ... 01000001 ^ Output: 10000010 Thus the printed output is in reverse. The simplest fix is to change the order of the loop. for (int i = 7; i >= 0;

I'm trying to parse a binary file format in Haskell (Apple's binary property list format), and one of the things required by the format is to treat sequences of bytes as either (a) unsigned 1-, 2-, or 4-byte integers; (b) signed 8-byte integers; (c) 32-bit float s; and (d) 64-bit double s. Converting sequences of bytes to unsigned integers is easy, and even dealing with signed integers wouldn't be terrible . But for signed integers, and especially for Float s and Double s, I don't really want to implement the logic myself. I've been able to find functions int2Float# :: Int# -> Float# and

This question already has answers here : How to count the number of set bits in a 32-bit integer? (53 answers) Closed 5 years ago . /*A value has even parity if it has an even number of 1 bits. *A value has an odd parity if it has an odd number of 1 bits. *For example, 0110 has even parity, and 1110 has odd parity. *Return 1 iff x has even parity. */ int has_even_parity(unsigned int x) { } I'm not sure where to begin writing this function, I'm thinking that I loop through the value as an array and apply xor operations on them. Would something like the following work? If not, what is the way to

I was asked the following question in an interview: int foofoo(unsigned int u) { unsigned int foo = u; do{ u = u/2; foo -= u; }while(u > 0); return foo; } I was asked to tell what does this function do and I was able to find that it counts the number of set bits in an unsigned int value, but I was not able to prove that, maybe someone can? I was able to find that it counts the number of set bits in an unsigned int value, but I was not able to prove that, maybe someone can? Look at a single bit m inside U . That bit contributes to the value of U like: U = ...... + U m 2^m + ..... (where the ...

I'm trying to write a function to return the number of bits a positive integer less that the Javascript limit of (2^53)-1 is. However im being hit by precision problems, and want to avoid big integer libraries. Method 1: function bitSize(num) { return Math.floor( Math.log(num) / Math.log(2) ) + 1; } Pass: bitSize( Math.pow(2, 16) -1 ) = 16 Pass: bitSize( Math.pow(2, 16) ) = 17 Fail (Should be 48): bitSize( Math.pow(2, 48) -1 ) = 49 Pass: bitSize( Math.pow(2, 48) ) = 49 Method 2: function bitSize(num) { var count = 0; while(num > 0) { num = num >> 1; count++; } return count; } Pass: bitSize(