c反向无符号整数位 [英] C reverse bits in unsigned integer

问题描述

我转换一个无符号整数使用位运算符为二进制，目前做的整数放; 1，检查是否位为1或0输出，然后通过1右移由2.然而比特错误的顺序（逆向）返回划分，所以我想扭转位开始前的整数订购。

I'm converting an unsigned integer to binary using bitwise operators, and currently do integer & 1 to check if bit is 1 or 0 and output, then right shift by 1 to divide by 2. However the bits are returned in the wrong order (reverse), so I thought to reverse the bits order in the integer before beginning.

有没有一种简单的方法来做到这一点？

Is there a simple way to do this?

例如：
所以，如果我给出的unsigned int类型10 = 1010

Example: So if I'm given the unsigned int 10 = 1010

while (x not eq 0) 
  if (x & 1)
    output a '1'
  else 
    output a '0'
  right shift x by 1

这将返回0101这是不正确......所以我想最初运行循环​​之前扭转位的顺序，但我不确定如何做到这一点？

this returns 0101 which is incorrect... so I was thinking to reverse the order of the bits originally before running the loop, but I'm unsure how to do this?

推荐答案

在一个字反转位是恼人的，它只是将它们输出以相反的顺序更容易。如，

Reversing the bits in a word is annoying and it's easier just to output them in reverse order. E.g.,

void write_u32(uint32_t x)
{
    int i;
    for (i = 0; i < 32; ++i)
        putchar((x & ((uint32_t) 1 << (31 - i)) ? '1' : '0');
}

下面是典型的解决方案，以扭转位顺序：

Here's the typical solution to reversing the bit order:

uint32_t reverse(uint32_t x)
{
    x = ((x >> 1) & 0x55555555u) | ((x & 0x55555555u) << 1);
    x = ((x >> 2) & 0x33333333u) | ((x & 0x33333333u) << 2);
    x = ((x >> 4) & 0x0f0f0f0fu) | ((x & 0x0f0f0f0fu) << 4);
    x = ((x >> 8) & 0x00ff00ffu) | ((x & 0x00ff00ffu) << 8);
    x = ((x >> 16) & 0xffffu) | ((x & 0xffffu) << 16);
    return x;
}

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