前一段时间在网上找以图搜图的资料，发现了感知哈希算法，得出的哈希值以字符串形式保存，我们管这种形式叫做哈希指纹。字符串中为十六进制数，需计算其二进制中的明氏距离，即二进制表示中不一样位数的个数。如下测试代码：

// test.cpp : 定义控制台应用程序的入口点。
//

#include "stdafx.h"
#include <iostream>
using namespace std;

int LSH(string str1,string str2);
int sub(char a,char b);

int _tmain(int argc, _TCHAR* argv[])
{
	string str1 = "9e0f53";
	string str2 = "0b672a";

	cout<<"str1 = "<<str1.c_str()<<endl;
	cout<<"str2 = "<<str2.c_str()<<endl;

	int dis = LSH(str1,str2);
	cout<<"dis = "<<dis<<endl;
	int a;
	cin >> a;
	return 0;
}

int LSH(string str1,string str2)
{
	//int L = sizeof(str1.c_str());
	int L = str1.length();
	cout<<"L = "<<L<<endl;
	int dis  = 0;
	for(int i=0;i<L;i++)
	{
		dis += sub(str1[i],str2[i]); 
	}
	return dis;
}

unsigned char char2hex(char a)
{
	unsigned char t(0);
	switch(a)
	{
	case 'a': t = 10; break;
	case 'b': t = 11; break;
	case 'c': t = 12; break;
	case 'd': t = 13; break;
	case 'e': t = 14; break;
	case 'f': t = 15; break;
	default: t = a - '0'; break;;
	}
	return t;
}

int sub(char a,char b)
{
	unsigned char a1 = char2hex(a);
	unsigned char b1 = char2hex(b);
	unsigned char c1 = a1^b1;
	cout<<"a1 = "<< (int)a1 <<endl;
	cout<<"b1 = "<< (int)b1 <<endl;
	//cout<<"c1 = "<< (int)c1 <<endl;

	int t(0);
	unsigned char tem(1);
	for(int i=0;i<8;i++)
	{
		if(c1&(tem))
			t++;
	   tem = tem<<1;
	}
	cout<<"t = "<<(int)t <<endl;
	return t;
}

运行结果：

<img src=http://www.mamicode.com/"http://img.blog.csdn.net/20140926140419752?watermark/2/text/aHR0cDovL2Jsb2cuY3Nkbi5uZXQveGF1dF96amI=/font/5a6L5L2T/fontsize/400/fill/I0JBQkFCMA==/dissolve/70/gravity/Center" style="font-family: Arial, Helvetica, sans-serif;" alt="" />

string型哈希值的明氏距离比较