Count number of bits in a 64-bit (long, big) integer?

问题:

I have read through this SO question about 32-bits, but what about 64-bit numbers? Should I just mask the upper and lower 4 bytes, perform the count on the 32-bits and then add them together?

回答1:

You can find 64 bit version here http://en.wikipedia.org/wiki/Hamming_weight

It is something like this

static long NumberOfSetBits(long i) {     i = i - ((i >> 1) & 0x5555555555555555);     i = (i & 0x3333333333333333) + ((i >> 2) & 0x3333333333333333);     return (((i + (i >> 4)) & 0xF0F0F0F0F0F0F0F) * 0x101010101010101) >> 56; } 

This is a 64 bit version of the code form here How to count the number of set bits in a 32-bit integer?

Using Joshua's suggestion I would transform it into this:

static int NumberOfSetBits(ulong i) {     i = i - ((i >> 1) & 0x5555555555555555UL);     i = (i & 0x3333333333333333UL) + ((i >> 2) & 0x3333333333333333UL);     return (int)(unchecked(((i + (i >> 4)) & 0xF0F0F0F0F0F0F0FUL) * 0x101010101010101UL) >> 56); } 

EDIT: I found a bug while testing 32 bit version. I added missing parentheses. The sum should be done before bitwise &, in the last line

EDIT2 Added safer version for ulong

回答2:

Standard answer in C#:

ulong val = //whatever byte count = 0;  while (val != 0) {     if ((val & 0x1) == 0x1) count++;     val >>= 1; } 

This shifts val right one bit, and increments count if the rightmost bit is set. This is a general algorithm that can be used for any length integer.

回答3:

A fast (and more portable than using non-standard compiler extensions) way:

int bitcout(long long n) {    int ret=0;    while (n!=0)    {        n&=(n-1);        ret++;    }    return ret; } 

Every time you do a n&=(n-1) you eliminate the last set bit in n. Thus this takes O(number of set bits) time.

This faster than the O(log n) you would need if you tested every bit - not every bit is set unless the number is 0xFFFFFFFFFFFFFFFF), thus usually you need far fewer iterations.

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