Question

我需要有关密码的帮助。我一切正常，直到遇到'a'后需要回到'z'的程序。我不确定该如何处理，因此任何输入都将有所帮助。为了清楚起见，我迷失的部分是if ('c' > z)所在的地方。

#include <stdio.h>
#include <stdlib.h>
#include <cs50.h>
#include <ctype.h>
#include <string.h>

// using argc and argv to get a shift number from the command line then 
// running a check to see the value of it
int main(int argc, string argv[])
{
char c;
int key = 0;

if (argc == 2)
{
    key = atoi(argv[1]);
}
else if (argc == 1)
{
    printf("ERROR\n");
    return 1;
}

string message = get_string("Enter a message to encrypt: ");
//getting the message you want to change
int len = strlen(message);

//loop to see when the array ends and run through it to
//shift each letter
for (int i = 0; message[i] != '\0'; i++)
{
    c = message[i];
//shifting the letters
    if (c >= 'a' && c <= 'z')
    {
        c = c + key;
//this is what doesn't work, when it hits z it doesn't go 
//back up and i'm not sure how to make it start at 'a' 
//again and then continue counting.
        if(c > 'z')
        {
            continue;
        }

        message[i] = c;
    }
//same as above with same error
    else if (c >= 'A' && c <= 'Z')
    {
        c = c + key;

        if (c > 'Z')
        {
            continue;
        }

        message[i] = c;
    }
}

printf("ciphertext: %s\n", message);
}

Answer 1

如果您希望将'z' + 1之类的值变为a，将'z' + 2之类的值变为b，依此类推，则可以更改此值：

    if(c > 'z')
    {
        continue;
    }

类似于：

    if(c > 'z')
    {
        c = c - 'z' + 'a' - 1;
    }

顺便说一句：您的代码无法处理溢出，因此不适用于任何key值。较小的键值会很好，但是某些较高的键值会引起问题。考虑以下代码：

#include <stdio.h>

int main(void) {
    char c = 'z';
    int key = 6;
    c = c + key;
    if ('c' > 'z')
    {
        printf("Greater than z\n");
    }
    else
    {
        printf("Less than z or equal to z\n");
    }

    int x  = c;
    printf("%d\n", x);
    return 0;
}

在某些平台（例如https://ideone.com/2LOvCU）上，这可能会提供输出

Less than z or equal to z
-128

我想这不是你想要的。也许您应该将所有计算都保留为（无符号）整数类型。

凯撒密码无法正确计算C代码

1 个答案: