C++二進制翻轉實例分析
2020-05-23 14:21:45
供稿：網友
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。
