C++二进制翻转实例分析

2020-01-26 15:17:53

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本文实例讲述了C++二进制翻转的方法，将常用的几种解决方法罗列出来供大家比较选择。具体如下：

首先来看看一个相对笨拙的算法：

#include <iostream>using namespace std;void printBinary(unsigned char str, int size = 1){ int flag = 0x01; for (int i = 0; i < size; i++) { for (int i = 0; i < 8; i++) {  if (str & (0x01 << (7 - i)))  cout << "1";  else  cout << "0"; } cout << endl;; }}unsigned char mySwap(unsigned char data){ unsigned char flag = 0x01; for (int i = 0, j = 7; i < j; i++, j--) { int right = data & (0x01 << i); int left = data & (0x01 << j); data &= ~(0x01 << j); data &= ~(0x01 << i); int dist = j - i; data |= (right << dist); data |= (left >> dist); } return data;}void main(void){ char source=0x07; int i; printBinary(source, 1); unsigned char result = mySwap(source); printBinary(result);}

下面这个翻转程序相对上面实例而言简洁高效：

unsigned char swapBinary(unsigned char data){ int sign = 1; unsigned char result = 0; for (int i = 0; i <= 7; i++) { result += ((data & (sign << i)) >> i) << (7 - i); } return result;}

下面这个反转程序比较容易理解：

unsigned char swapBinary2(unsigned char data){ data=(( data & 0xf0) >> 4) | ((data & 0x0f) << 4);  data=((data & 0xCC) >> 2) | ((data & 0x33) << 2);  data=((data & 0xAA) >> 1) | ((data & 0x55) << 1);  return data; }

最后这个超牛的反转程序简直碉堡了。。。

unsigned char codeTable[16]={0x00, 0x08, 0x04, 0x0c, 0x02, 0x0a, 0x06, 0x0e, 0x01, 0x09, 0x05, 0x0d, 0x03, 0x0b, 0x07, 0x0f}; unsigned char swapBinary3(unsigned char data) {  return ((codeTable[data >> 4]) | (codeTable[data & 0x0f] << 4));}

希望本文所述对大家C++程序算法设计的学习有所帮助。

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