Question

我正在尝试从哈希中破解最多4个字母的密码。理想情况下，我可以写./crack 50fkUxYHbnXGw并返回rofl。

我的方法是使用嵌套的for循环。为了将它应用于不同长度的字符串，我建议在C中使用空终止符，但是现在我已经尝试了每个可能长度的if语句。

你知道更优雅的方式吗？

最后，我有一个错误，它没有给我任何输出。

#define _XOPEN_SOURCE // for crypt
#include <cs50.h> // used for get_string
#include <stdio.h> // used for printf
#include <string.h> // used for strlen
#include <ctype.h> // used for isalpha
#include <stdlib.h> // for atoi
#include <crypt.h> // for crypt

int main(int argc, char* argv[])
{
    // accept only two arguments from command line
    // hash is max 13 characters. First two are the salt.
    // next 11 are the key
    if (argc != 2)
    {
        printf("Usage: ./crack hash \n");
        return 1;
    }

    // make second command line argument into string
    char* hash = argv[1];
    char* salt = argv[1];

    strncpy(salt, hash, 2);
    salt[2] = 0; // null terminates destination

    char* key = hash + 2; // just key without salt

    char* abAB = "abcdefghijklmnopqrstuvwxyzQWERTYUIOPASDFGHJKLZXCVBNM";

    for (int i = 0, n = strlen(abAB); i < n; i++)
    {   //check for every length
        char tri1[1] = {abAB[i]};
        // produce my own encrpyted key with salt
        char* cr1 = crypt(tri1, salt);
        // if own key is equal to key provided, print out the string tried.
        int comp1 = strcmp(cr1, key);

        if (comp1 == 0)
        {
            printf("%s", tri1);
        }

        for (int j = 0, m = strlen(abAB); j < m; j++)
        {
                char tri2[2] = {abAB[i],abAB[j]};
                char* cr2 = crypt(tri2, salt);
                int comp2 = strcmp(cr2, key);
                if (comp2 == 0)
                {
                    printf("%s", tri2);
                }

            for (int k = 0, p = strlen(abAB); k < p; k++)
            {
                char tri3[3] = {abAB[i],abAB[j],abAB[k]};
                char* cr3 = crypt(tri3, salt);
                int comp3 = strcmp(cr3, key);
                if (comp3 == 0)
                {
                    printf("%s", tri3);
                }

                for (int l = 0, q = strlen(abAB); l < q; l++)
                {
                    char tri4[4] = {abAB[i],abAB[j],abAB[k],abAB[l]};
                    // produce my own encrpyted key with salt
                    char* cr4 = crypt(tri4, salt);
                    // if own key is equal to key provided, print out the string tried.
                    int comp4 = strcmp(cr4, key);
                    if (comp4 == 0)
                    {
                        printf("%s", tri4);
                    }
                }
            }
        }
    }
    printf("\n");
    return 0;
}

Answer 1

这是我最后对那些感兴趣的人所做的。我还不确定如何以较少的重复来改变琴弦的长度。

#define _XOPEN_SOURCE // for crypt
#include <cs50.h> // used for get_string
#include <stdio.h> // used for printf
#include <string.h> // used for strlen
#include <ctype.h> // used for isalpha
#include <stdlib.h> // for atoi
#include <crypt.h> // for crypt

int main(int argc, char* argv[])
{

// accept only two arguments from command line
// hash is max 13 characters. First two elements are the salt,
// the other 11 are the key

if (argc != 2)
{
    printf("Usage: ./crack hash \n");
    return 1;
}

// make second command line argument into string
char* hash = argv[1];
char salt[3];
strncpy(salt, hash, 2);
salt[2] = '\0'; // null terminates destination

char* key = hash + 2; // just key without salt

printf("key: %s\n", key);
printf("salt: %s\n", salt);

char* abAB = "abcdefghijklmnopqrstuvwxyzQWERTYUIOPASDFGHJKLZXCVBNM";

for (int i = 0, n = strlen(abAB); i < n; i++)
{
    //check for every length
    char tri1[2] = {abAB[i], 0};

    // produce my own encrpyted key with salt
    char* cr1 = crypt(tri1, salt);

    // if own key is equal to key provided, print out the string tried.
    if (strcmp(cr1, hash) == 0)
    {
        printf("%s\n", tri1);
        return 0;
    }

    for (int j = 0, m = strlen(abAB); j < m; j++)
    {
        char tri2[3] = {abAB[i], abAB[j], 0};
        char* cr2 = crypt(tri2, salt);

        if (strcmp(cr2, hash) == 0)
        {
            printf("%s\n", tri2);
            return 0;
        }

        // this didn't work - do you know why???
        // int comp2 = strcmp(cr2, hash);
        // if (comp2 == 0)
        // {
        //     printf("%s", tri2);
        //     printf("test");
        //     return 0;
        // }

        for (int k = 0, p = strlen(abAB); k < p; k++)
        {
            char tri3[4] = {abAB[i],abAB[j],abAB[k], 0};
            char* cr3 = crypt(tri3, salt);

            if (strcmp(cr3, hash) == 0)
            {
                printf("%s\n", tri3);
                return 0;
            }

            for (int l = 0, q = strlen(abAB); l < q; l++)
            {
                char tri4[5] = {abAB[i],abAB[j],abAB[k],abAB[l], 0};
                // produce my own encrpyted key with salt
                char* cr4 = crypt(tri4, salt);

                // if own key is equal to key provided, print out the string tried.
                if (strcmp(cr4, hash) == 0)
                {
                    printf("%s\n", tri4);
                    return 0;
                }
            }
        }
    }
}

printf("\n");
return 1;
}

用暴力破解短密码：嵌套for循环

1 个答案: