按位反转数据 Write a C function to swap the bits of a unsigned char so that its bits become the mirror image of the char. MSBs become its LSBs, e.g. 01111000 binary should become 00011110 binary.
先按照题目的要求来看看：
1. 简单而容易想到的做法是，将前后对应的位置对调；
unsigned char reverseBits(unsigned char ch)
{
int bitLen =sizeof(ch);
unsigned char leftBit =0x80;
unsigned char rightBit =0x1;
unsigned char temp=0; //必须要赋初始值
for(int i=0; i<bitLen/2; i++)
{
temp |=((ch & leftBit)>>(bitLen-1-i*2));
temp |=((ch & rightBit)<<(bitLen-1-i*2));
leftBit >>=1;
rightBit <<=1;
}
return temp;
}
做这个处理有一个经典的方法，分治策略( divide and conquer strategy )，该方法来源于Henry S.Warren 的《Hacker's Delight》对Reversing Bits 的介绍。
unsigned char ReverseBits(unsigned char ch)
{
ch = (ch & 0x55) << 1 | (ch >> 1) & 0x55;
ch = (ch & 0x33) << 2 | (ch >> 2) & 0x33;
ch = (ch & 0x0F) << 4 | (ch >> 4) & 0x0F;
return ch;
}
先看看0x55，0x33，0x0F的二进制
0x55 -> 01010101 B
0x33 -> 00110011 B
0x0F -> 00001111 B
从中可以看出，是先将相连的两bits 对调－－实现相连2 bits数据翻转；接着两个相连的“2 bits组合”对调－－实现相连4 bits数据翻转；再下来就是将两个相连的“4 bits组合”对调－－即可以完成对一个字符8 bits的翻转。对于更多bytes 的数据做同样的处理。单数个bits呢，这个就需要思考一下了。
至于多bytes 的bits 翻转可以写一些上面函数的重载函数；
写一个翻转int ，short，char 类型data 的函数，感觉还不如重载来的方便；不过既然写了，就放上来；
#include <cassert>
#include <iostream>
using namespace std;
typedef struct Datatype {
union Dtype
{
unsigned int idata;
unsigned short sdata;
unsigned char cdata;
} instndtype;
int size;
} datatype;
int pow2(int pp) {
int result=1;
while (pp--) {
result *=2;
}
return result;
}
int log22(int pp) {
assert(pp>=1);
int result =0;
while (pp/=2) {
result++;
}
return result;
}
Datatype ReverseBits(Datatype ch)
{
unsigned char* pchdata =NULL;
unsigned short* psdata =NULL;
unsigned int* pidata =NULL;
int i=0;
switch(ch.size) {
case 1:
pchdata =new unsigned char[3];
*pchdata =0x55, *(pchdata+1) =0x33, *(pchdata+2) =0x0F;
for (i =0; i<3; i++ )
{
ch.instndtype.cdata = (ch.instndtype.cdata & *(pchdata+i)) <<pow2(i)
| (ch.instndtype.cdata >>pow2(i) & *(pchdata+i));
}
delete []pchdata;
break;
case 2:
psdata =new unsigned short[4];
*psdata =0x5555, *(psdata+1) =0x3333, *(psdata+2) =0x0F0F, *(psdata+3) =0x00FF;
for (i =0; i<4; i++ )
{
ch.instndtype.sdata = (ch.instndtype.sdata & *(psdata+i)) <<pow2(i)
| (ch.instndtype.sdata >>pow2(i) & *(psdata+i));
}
delete []psdata;
psdata =NULL;
break;
case 4:
pidata =new unsigned int[5];
*pidata =0x55555555, *(pidata+1) =0x33333333, *(pidata+2) =0x0F0F0F0F,
*(pidata+3) =0x00FF00FF, *(pidata+4) =0x0000FFFF;
for (i =0; i<5; i++ )
{
ch.instndtype.idata = (ch.instndtype.idata & *(pidata+i)) <<pow2(i)
| (ch.instndtype.idata >> pow2(i) & *(pidata+i));
}
delete []pidata;
pidata =NULL;
break;
default:
;
}
return ch;
} //符号优先级 '&' > '|' ???
//调用
datatype numTest ={0x11FF, sizeof(short)};
unsigned short chResult =ReverseBits(numTest).instndtype.sdata;
内容来源：
http://blog.csdn.net/ammana_babi/archive/2007/06/21/1660983.aspx