Question

所以我正在学习计算机科学的edX课程之一（不是为了学分，只是为了帮助自学），高级问题集与解密密码有关。为了更好地理解用于进行加密的函数crypt（），我做了以下代码来弄清楚用于“盐”字符串的内容（终端中的man crypt（）命令说:(“salt是一个从字符集[a-zA-Z0-9./]中选择的两个字符的字符串。这个字符串用于以4096种不同的方式之一扰乱算法。“））请告诉我你是否可以理解为什么我是得到这个错误。

以下是我使用的代码。我在尝试编译时收到此错误：

test.c:37:33: error: incompatible integer to pointer conversion passing 'char' to
parameter of type 'char *'; take the address with & [-Werror,-Wint-conversion]
char *salt = strcat(whatever1,whatever2);

代码：

    #define _XOPEN_SOURCE //this is needed for the unistd and crypt stuff
    #include <stdio.h>
    #include <unistd.h> //this lets me do the crypt() fcn
    #include <cs50.h> //just lets me use string as a type among w/some input fcns and       is made by the course
    #include <string.h>
    #include <ctype.h> //this lets me use toascii()

    char *word = "caesar";
    char *password;

    char toASCII_New(int i) //attempt to turn input to HEX then corresponding ASCII character
    {
        for(int num = i; num<= 172; num++)
        {
            for(int hex_num = 0x41; hex_num <= 0x7a; hex_num++)
            {
                if(num == hex_num)
                {
                    char letter = (char) toascii(hex_num); //should convert hex to ascii here
                    return letter;
                }
            }
        }
        char backup = 'a';
        return backup;
    }


    string checker(int first, int second) //Checks the inputs to see if the can be the correct SALTS input to crypt
    {
        for (int i = first; i<=first+25; i++)
        {
            for(int j = second; j<=second+25; j++)
            {
                char whatever1 = (char) toASCII_New(i);
                char whatever2 = (char) toASCII_New(j);
                char *salt = strcat(whatever1,whatever2);
                password = crypt(word,salt);
                int compare = strcmp(password,"50zPJlUFIYY0o"); // this is what i'm really checking
                if (compare == 0)
                    return salt;

            }
        }
    return "no";
    }


    int main(void)
    {
        string AA = checker(65,65);
        string Aa = checker(65,97);
        string aA = checker(97,65);
        string aa = checker(97,97);

        printf("AA: %s\nAa: %s\naA: %s\naa: %s\n",AA,Aa,aA,aa);
    }

Answer 1

strcat用于连接字符串而不是2个单独的字符。＆＃34; whatever1＆＃34;将连接的字符串保持在一起。为什么要使用-dangerous-strcat操作，只需要一个2字节的字符串？只需使用：

char salt[3];
salt[0]=toASCII_New(i); /* why the casting to char ? you already returned a char... */
salt[1]=toASCII_New(j);
salt[2]='\0'; /* to terminate the string properly */

Answer 2

以下是一些一般性建议，因为整个代码需要重构：

函数toASCII_New应该按如下方式重写：

char toASCII_New(int c)
{
    if (('A' <= c && c <= 'Z') || ('a' <= c && c <= 'z'))
        return (char)toascii(c);
    return 'a';
}

函数checker应该按如下方式重写：

int checker(int first, int second, char salt[3])
{
    for (int i=first; i<=first+25; i++)
    {
        for (int j=second; j<=second+25; j++)
        {
            salt[0] = toASCII_New(i);
            salt[1] = toASCII_New(j);
            password = crypt(word,salt);
            if (strcmp(password,"50zPJlUFIYY0o") == 0)
                return 1;
        }
    }
    return 0;
}

函数main应该按如下方式重写：

int main()
{
    char salt_AA[3] = {0};
    char salt_Aa[3] = {0};
    char salt_aA[3] = {0};
    char salt_aa[3] = {0};
    int  salt_AA_OK = checker('A','A');
    int  salt_Aa_OK = checker('A','a');
    int  salt_aA_OK = checker('a','A');
    int  salt_aa_OK = checker('a','a');
    printf("AA: %s\n",salt_AA_OK? salt_AA:"failed");
    printf("Aa: %s\n",salt_Aa_OK? salt_Aa:"failed");
    printf("aA: %s\n",salt_aA_OK? salt_aA:"failed");
    printf("aa: %s\n",salt_aa_OK? salt_aa:"failed");
    return 0;
}

C编译器认为char类型是int类型 - clang编译器

2 个答案: