C语言写的MD5加密程序重复执行问题

[olindung]

在main()函数尾部重复调用自己，编译运行，只有第一次加密得出了正确结果，那位大牛能解释一下，谢谢！
#include<iostream>
#include<string.h>
#include<stdio.h>
#define MAXNUM 65536

using namespace std;

char info[MAXNUM]="\0";
unsigned long int num=0;
unsigned long A=0x67452301;
unsigned long B=0xefcdab89;
unsigned long C=0x98badcfe;
unsigned long D=0x10325476;

int F(int X,int Y,int Z)
{
return (X&Y)|((~X)&Z);
}
int G(int X,int Y,int Z)
{
return (X&Z)|(Y&(~Z));
}
int H(int X,int Y,int Z)
{
return X^Y^Z;
}
int I(int X,int Y,int Z)
{
return Y^(X|(~Z));
}


void FF(unsigned long &a,unsigned long b,unsigned long c,unsigned long d,unsigned long M,unsigned long s,unsigned long ti)
{
unsigned long tmp=(a+F(b,c,d)+M+ti);
a=(tmp<<s)|(tmp>>(32-s));
a=a+b;
}
void GG(unsigned long &a,unsigned long b,unsigned long c,unsigned long d,unsigned long M,unsigned long s,unsigned long ti)
{
unsigned long tmp=(a+G(b,c,d)+M+ti);
a=(tmp<<s)|(tmp>>(32-s));
a=a+b;
}
void HH(unsigned long &a,unsigned long b,unsigned long c,unsigned long d,unsigned long M,unsigned long s,unsigned long ti)
{
unsigned long tmp=(a+H(b,c,d)+M+ti);
a=(tmp<<s)|(tmp>>(32-s));
a=a+b;
}
void II(unsigned long &a,unsigned long b,unsigned long c,unsigned long d,unsigned long M,unsigned long s,unsigned long ti)
{
unsigned long tmp=(a+I(b,c,d)+M+ti);
a=(tmp<<s)|(tmp>>(32-s));
a=a+b;
}

void input()
{
printf("Please enter the origin information:\n");
gets(info);
//char str1[5]={'a','b','c','d'};
//info= str1
num=strlen(info);
}

void append()
{
unsigned long int tmp=num*8;
int i;
for(i=0;i<64;i++)
if((num+i)%64==56)
break;
if(i)
{
info[num++]=(char)0x80;
while(--i)
info[num++]=0;
}
memmove(info+num,(char*) &tmp,sizeof(tmp));
num+=8;
}

int main()
{
int i;
unsigned long a;
unsigned long b;
unsigned long c;
unsigned long d;
unsigned long M[16];

input();
append();
for(i=0;i<num;i=i+64)
{
a=A;b=B;c=C;d=D;
memmove((char*)M,info+i,64);
//第一轮
FF(a, b, c, d, M[0], 7, 0xd76aa478);
FF(d, a, b, c, M[1], 12, 0xe8c7b756);
FF(c, d, a, b, M[2], 17, 0x242070db);
FF(b, c, d, a, M[3], 22, 0xc1bdceee);
FF(a, b, c, d, M[4], 7, 0xf57c0faf);
FF(d, a, b, c, M[5], 12, 0x4787c62a);
FF(c, d, a, b, M[6], 17, 0xa8304613);
FF(b, c, d, a, M[7], 22, 0xfd469501);
FF(a, b, c, d, M[8], 7, 0x698098d8);
FF(d, a, b, c, M[9], 12, 0x8b44f7af);
FF(c, d, a, b, M[10], 17, 0xffff5bb1);
FF(b, c, d, a, M[11], 22, 0x895cd7be);
FF(a, b, c, d, M[12], 7, 0x6b901122);
FF(d, a, b, c, M[13], 12, 0xfd987193);
FF(c, d, a, b, M[14], 17, 0xa679438e);
FF(b, c, d, a, M[15], 22, 0x49b40821);
//第二轮
GG(a, b, c, d, M[1], 5, 0xf61e2562);
GG(d, a, b, c, M[6], 9, 0xc040b340);
GG(c, d, a, b, M[11], 14, 0x265e5a51);
GG(b, c, d, a, M[0], 20, 0xe9b6c7aa);
GG(a, b, c, d, M[5], 5, 0xd62f105d);
GG(d, a, b, c, M[10], 9, 0x02441453);
GG(c, d, a, b, M[15], 14, 0xd8a1e681);
GG(b, c, d, a, M[4], 20, 0xe7d3fbc8);
GG(a, b, c, d, M[9], 5, 0x21e1cde6);
GG(d, a, b, c, M[14], 9, 0xc33707d6);
GG(c, d, a, b, M[3], 14, 0xf4d50d87);
GG(b, c, d, a, M[8], 20, 0x455a14ed);
GG(a, b, c, d, M[13], 5, 0xa9e3e905);
GG(d, a, b, c, M[2], 9, 0xfcefa3f8);
GG(c, d, a, b, M[7], 14, 0x676f02d9);
GG(b, c, d, a, M[12], 20, 0x8d2a4c8a);
//第三轮
HH(a, b, c, d, M[5], 4, 0xfffa3942);
HH(d, a, b, c, M[8], 11, 0x8771f681);
HH(c, d, a, b, M[11], 16, 0x6d9d6122);
HH(b, c, d, a, M[14], 23, 0xfde5380c);
HH(a, b, c, d, M[1], 4, 0xa4beea44);
HH(d, a, b, c, M[4], 11, 0x4bdecfa9);
HH(c, d, a, b, M[7], 16, 0xf6bb4b60);
HH(b, c, d, a, M[10], 23, 0xbebfbc70);
HH(a, b, c, d, M[13], 4, 0x289b7ec6);
HH(d, a, b, c, M[0], 11, 0xeaa127fa);
HH(c, d, a, b, M[3], 16, 0xd4ef3085);
HH(b, c, d, a, M[6], 23, 0x04881d05);
HH(a, b, c, d, M[9], 4, 0xd9d4d039);
HH(d, a, b, c, M[12], 11, 0xe6db99e5);
HH(c, d, a, b, M[15], 16, 0x1fa27cf8);
HH(b, c, d, a, M[2], 23, 0xc4ac5665);
//第四轮
II(a, b, c, d, M[0], 6, 0xf4292244);
II(d, a, b, c, M[7], 10, 0x432aff97);
II(c, d, a, b, M[14], 15, 0xab9423a7);
II(b, c, d, a, M[5], 21, 0xfc93a039);
II(a, b, c, d, M[12], 6, 0x655b59c3);
II(d, a, b, c, M[3], 10, 0x8f0ccc92);
II(c, d, a, b, M[10], 15, 0xffeff47d);
II(b, c, d, a, M[1], 21, 0x85845dd1);
II(a, b, c, d, M[8], 6, 0x6fa87e4f);
II(d, a, b, c, M[15], 10, 0xfe2ce6e0);
II(c, d, a, b, M[6], 15, 0xa3014314);
II(b, c, d, a, M[13], 21, 0x4e0811a1);
II(a, b, c, d, M[4], 6, 0xf7537e82);
II(d, a, b, c, M[11], 10, 0xbd3af235);
II(c, d, a, b, M[2], 15, 0x2ad7d2bb);
II(b, c, d, a, M[9], 21, 0xeb86d391);
A+=a;B+=b;C+=c;D+=d;
}
unsigned char r[16];
memmove(r+0,(char *)&A,4);
memmove(r+4,(char *)&B,4);
memmove(r+8,(char *)&C,4);
memmove(r+12,(char *)&D,4);
printf("MD5:\n");
char s1[32]="\0";
sprintf(s1,"%02x%02x%02x%02x%02x%02x%02x%02x%02x%02x%02x%02x%02x%02x%02x%02x",r[0],r[1],r[2],r[3],r[4],r[5],r[6],r[7],r[8],r[9],r[10],r[11],r[12],r[13],r[14],r[15]);

printf("%s\n",s1);
main();
}

[olindung]

[huanmie_09]

因为在main函数里你修改了全局变量A,B,C,D：
A+=a;B+=b;C+=c;D+=d;
这样导致下一轮递归的时候，A,B,C,D的初值发生了改变，从而导致加密有误.

修改方法：
在main函数显式给A,B,C,D赋值。
如：
int main()
{
int i;
unsigned long a;
unsigned long b;
unsigned long c;
unsigned long d;
unsigned long M[16];

A=0x67452301; //ADD
B=0xefcdab89;
C=0x98badcfe;
D=0x10325476;

[olindung]

谢谢，是这样的