C++获取位操作

起因

     获取某个值的指定几位(bit)

     例如:

          整数89它的二进制是1011001。这时我需要获取从该二进制从右开始第3位到5位的二进制,并转换位10进制。那么如何操作呢?有人说可以按位左移并右移得到,ok 理论这完全可以,但你可以试试,会出错!因为整数89,你并没有告诉系统这个值是什么类型,所以系统按照4字节处理时,你就不能用程序来控制移动了。ok,这里我们给出验证过的C++程序。

程序

#include "iostream"
#include "stdlib.h"

// 字符串转int
#define STR2INT(value) atoi(value)

#define DEBUG 0

unsigned int _bit_(const unsigned int &value,const unsigned short &start,const unsigned short &length)
{
    
    
	// 计算位长
	unsigned short bitCnt = 32;
	unsigned int Y = 0xFFFFFFFF;
	if(value <= 255)
	{
    
    
		bitCnt = 8;
		
		Y = 0xFF;
	}
	else if(value <= 65535)
	{
    
    
		bitCnt = 16;

		Y = 0xFFFF;
	}

#if DEBUG
	std::cout << "bitCnt:" << bitCnt << std::endl;
#endif

	// 计算左移步长
	unsigned short moveCnt = bitCnt - (start + length);

#if DEBUG
	std::cout << "moveCnt1:" << moveCnt << std::endl;
#endif

	// 开始左移
	unsigned int tmp = 0;
	tmp = (value << moveCnt);

#if DEBUG
	std::cout << "tmp1:" << tmp << std::endl;
#endif

	// 开始&
	tmp = (tmp & Y);

#if DEBUG
	std::cout << "tmp2:" << tmp << std::endl;
#endif

	// 计算右移步长
	moveCnt = moveCnt + start;

#if DEBUG
	std::cout << "moveCnt2:" << moveCnt << std::endl;
#endif

	// 开始右移
	tmp = (tmp >> moveCnt);

#if DEBUG
	std::cout << "tmp3:" << tmp << std::endl;
#endif

	return tmp;	
}

int main(int argc,char **argv)
{
    
    
#if 0
	if(argc == 4)
	{
    
    

		std::cout << _bit_(STR2INT(argv[1]),STR2INT(argv[2]),STR2INT(argv[3])) << std::endl;
	}
#else
	std::cout << _bit_(89,1,5) << std::endl;				// 12
#endif

/*
	unsigned short value = 89;

	std::cout << (value << 2) << std::endl;

	tmp = (value << 2);

	tmp = tmp & 0xFF;

	std::cout << tmp << std::endl;

	tmp = (tmp >> 3);

	std::cout << tmp << std::endl;
*/

	
	return 0;
}

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笔者 - jxd

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转载自blog.csdn.net/automoblie0/article/details/107847715