我想在文件中编写一个名为bitCount()
的函数:bitcount.c
,它返回其无符号整数参数的二进制表示中的位数。
这是我到目前为止所做的:
#include <stdio.h>
int bitCount (unsigned int n);
int main () {
printf ("# 1-bits in base 2 representation of %u = %d, should be 0\n",
0, bitCount (0));
printf ("# 1-bits in base 2 representation of %u = %d, should be 1\n",
1, bitCount (1));
printf ("# 1-bits in base 2 representation of %u = %d, should be 16\n",
2863311530u, bitCount (2863311530u));
printf ("# 1-bits in base 2 representation of %u = %d, should be 1\n",
536870912, bitCount (536870912));
printf ("# 1-bits in base 2 representation of %u = %d, should be 32\n",
4294967295u, bitCount (4294967295u));
return 0;
}
int bitCount (unsigned int n) {
/* your code here */
}
好的,当我刚刚运行时,我得到了:
# 1-bits in base 2 representation of 0 = 1, should be 0
# 1-bits in base 2 representation of 1 = 56, should be 1
# 1-bits in base 2 representation of 2863311530 = 57, should be 16
# 1-bits in base 2 representation of 536870912 = 67, should be 1
# 1-bits in base 2 representation of 4294967295 = 65, should be 32
RUN SUCCESSFUL (total time: 14ms)
它不会返回正确的位数。
在C中返回其无符号整数参数的二进制表示中的位数的最佳方法是什么?
答案 0 :(得分:7)
这是一个不需要迭代的解决方案。它利用了以二进制方式添加位完全独立于位的位置并且总和不超过2位的事实。 00+00=00
,00+01=01
,01+00=01
,01+01=10
。第一个添加同时添加16个不同的1位值,第二个添加8个2位值,后面每个值增加一半,直到只剩下一个值。
int bitCount(unsigned int n)
{
n = ((0xaaaaaaaa & n) >> 1) + (0x55555555 & n);
n = ((0xcccccccc & n) >> 2) + (0x33333333 & n);
n = ((0xf0f0f0f0 & n) >> 4) + (0x0f0f0f0f & n);
n = ((0xff00ff00 & n) >> 8) + (0x00ff00ff & n);
n = ((0xffff0000 & n) >> 16) + (0x0000ffff & n);
return n;
}
这是硬编码为32位整数,如果你的大小不同,则需要进行调整。
答案 1 :(得分:6)
int bitCount(unsigned int n) {
int counter = 0;
while(n) {
counter += n % 2;
n >>= 1;
}
return counter;
}
答案 2 :(得分:4)
事实证明,有一些非常复杂的方法可以按照here的答案进行计算。
下面的impl(我学会了回过头来)只是在每次迭代时循环敲掉最不重要的位。
int bitCount(unsigned int n) {
int counter = 0;
while(n) {
counter ++;
n &= (n - 1);
}
return counter;
}