assembly - 汇编代码。加密到解密例程

Question

• 有人可以帮我看看这个并帮助我计算出一个可以反转用户输入的字符串的解密方法。我并不是说只是做这个过程的相反。

    push edx 
    push ecx 
    not eax 
    add eax,0x04 
    mov edx,eax 
    pop eax 
    xor eax,edx 
    pop edx 
    rol al,1 
    rol al,1
    rol al,1 
    sub al,0x02 
    ret
  

  *

寄存器是： Inwards- ecx：加密密钥。eax：要加密的字符。

Outwards-eax：加密字符

感谢您花时间查看。

Answer 1

该算法是对称的，因为我可以解密每个字符和密钥组合。

这两个函数通过一个循环来测试在解密返回值时是否有任何错误：

#include <iostream>
using namespace std;

unsigned char enc(unsigned char ch, unsigned char key)
{
    unsigned char tmp = key^(~ch+(unsigned char)0x04);
    return (( (tmp<<3) | (tmp>>5) ) & 0xff)-0x02;
}

unsigned char dec(unsigned char ch, unsigned char key)
{
    unsigned char tmp = (ch+0x02);
    tmp = ((tmp>>3) | (tmp<<5)) & 0xff;
    return ~((tmp^key )-(unsigned char)0x04);
}

int main()
{
    // single encryption test
    char c = 'A';
    char key = 'K';
    cout << "single test:" << (char)enc(c, key) << endl;

    bool problem = false;
    int k, ch;
    for(k=0;k<256;k++)
    {

        for(ch=0;ch<256;ch++)
        {
            if( enc( dec((unsigned char)ch, (unsigned char)k), k) != ch )
            {
                problem = true;
                cout << "error k=" << k << "c=" << ch
                     << "result=" <<  (unsigned int)enc( dec((unsigned char)ch, (unsigned char)key), (unsigned char)key) << endl;

            }
        }
    }
    if(problem) cout << "There was a problem." << endl;
    else cout << "There was no problem." << endl;
}