實現：
//00000000000000000000000000011001
//10011000000000000000000000000000
這種翻轉
方法一和方法二思路大致一樣，可以方法二卻十分麻煩。
方法一（簡單）
#include<stdio.h>
#include<math.h>
unsigned int reverse_bit(unsigned int value)
{
int i = 0;
unsigned int Value ;
unsigned int num = 0;
for (i = 1; i <= 32; i++)
{
Value = value;
int b = ((value%2)&&1);
num = num + b*(pow(2, (32 - i)));
value=Value >> 1;
}
return num;
}
int main()
{
unsigned int number = 25;
printf("%u", reverse_bit(number));
}
方法二（思想簡單但寫的十分複雜，我都不知道我咋想出來的）#include<stdio.h>
#include<math.h>
unsigned int zhishu(int x, int cishu)
{
unsigned int num = pow(2, cishu);
return (x*num);
}
int num_sing(int max)
{
if (max % 2 == 0)
{
return 0;
}
else
{
return 1;
}
}
unsigned int fzsign(unsigned int a)
{
unsigned int num = 0;
int sing = num_sing(a);
int cishu = 0 + sing;
unsigned int tmp = a;
unsigned int b = 0;
while (a != sing)
{
a = a / 2;
cishu++;
}
/*printf("%d\n", cishu);*/
int sings = num_sing(tmp);
int tmp2 = cishu;
cishu = cishu - 1;
while (tmp != sings)
{
b = tmp % 2;
tmp = tmp / 2;
num = num + zhishu(b, cishu);
cishu--;
}
num = num + sings;
num = num << (32 - tmp2);
return num;
}
int main()
{
unsigned int a = 25;
unsigned int num=fzsign(a);
printf("%u", num);
}
法二像不像大雜燴！！！本文出自 “痕跡” 博客，請務必保留此出處http://wpfbcr.blog.51cto.com/10696766/1703666