设置的位数（可以设置的位数迭代次数相同）

Question

  

可能重复：
  Best algorithm to count the number of set bits in a 32-bit integer?
  Counting the number of bits that are set

如果允许迭代循环的次数与设置的位数相同，则如何找到无符号整数中设置的位数（如果设置了3位，则只能迭代循环3次）< / p>

Answer 1

这是我所知道的这样做的方法，它只会迭代v中设置的位数，在执行结束时c将保存v中的位数1}}：

unsigned int v; // count the number of bits set in v
unsigned int c; // c accumulates the total bits set in v
for (c = 0; v; c++)
{
  v &= v - 1; // clear the least significant bit set
}

来源：C编程语言第二版。 （作者：Brian W. Kernighan和Dennis M. Ritchie）

示例：

v = 1010; //2 bits are set
v - 1 = 1010(10 decimal) - 1 = 1001(9 decimal)

1010
1001 &
------
1000         The least significant bit is unset, c is incremented by 1

v = 1000 
v - 1 = 1000 - 1 = 0111

1000
0111 &
-------
0000         The least significant bit is unset again, c is incremented 
             by 1, the loop stops because there are no more set bits.