The best method for counting bits in a 32-bit integer-优快云博客

本文介绍了一种高效的32位整数中1的位数计数方法。通过巧妙利用位操作和预定义的常数值，该算法能在极短的时间内完成计算。示例展示了每一步的具体实现细节。

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B[0] = 0x55555555 = 01010101 01010101 01010101 01010101
B[1] = 0x33333333 = 00110011 00110011 00110011 00110011
B[2] = 0x0F0F0F0F = 00001111 00001111 00001111 00001111
B[3] = 0x00FF00FF = 00000000 11111111 00000000 11111111
B[4] = 0x0000FFFF = 00000000 00000000 11111111 11111111

The best method for counting bits in a 32-bit integer v is the following:

v = v - ((v >> 1) & 0x55555555);                    // reuse input as temporary
v = (v & 0x33333333) + ((v >> 2) & 0x33333333);     // temp
c = ((v + (v >> 4) & 0xF0F0F0F) * 0x1010101) >> 24; // count

http://graphics.stanford.edu/~seander/bithacks.html#CountBitsSetParallel


B[0] = 0x55555555 = 01010101 01010101 01010101 01010101
B[1] = 0x33333333 = 00110011 00110011 00110011 00110011
B[2] = 0x0F0F0F0F = 00001111 00001111 00001111 00001111
B[3] = 0x00FF00FF = 00000000 11111111 00000000 11111111
B[4] = 0x0000FFFF = 00000000 00000000 11111111 11111111

The best method for counting bits in a 32-bit integer v is the following:

v = v - ((v >> 1) & 0x55555555); // reuse input as temporary
v = (v & 0x33333333) + ((v >> 2) & 0x33333333); // temp
c = ((v + (v >> 4) & 0xF0F0F0F) * 0x1010101) >> 24; // count				1
	01010101	01010101	01010101	01010101
	v=			00000111
v >> 1				00000011
((v >> 1) & 0x55555555)	00000000	00000000	00000000	00000001
	11111111	11111111	11111111	11111111
v = v - ((v >> 1) & 0x55555555);	10000000	00000000	00000000	00000110
	00110011	00110011	00110011	00110011
(v & 0x33333333)	00000000	00000000	00000000	00000010
((v >> 2) & 0x33333333);				00000001
				00000011	0x1010101
					00000001000000010000000100000001
					00000001000000010000000100000001
c = ((v + (v >> 4) & 0xF0F0F0F) * 0x1010101) >> 24; // count					00000011