本文實例講述了C++二進制翻轉的方法，將常用的幾種解決方法羅列出來供大家比較選擇。具體如下：
首先來看看一個相對笨拙的算法：
#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);}下面這個翻轉程序相對上面實例而言簡潔高效：
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;}下面這個反轉程序比較容易理解：
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; }最后這個超牛的反轉程序簡直碉堡了。。。
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));}希望本文所述對大家C++程序算法設計的學習有所幫助。
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