Question

因此，我需要编写一个程序来加密将x（{argv[1] = x）添加到提示用户的文本上的文本。即：

./program 1  //running the program with argv[1] = 1

plaintext: abcd  //prompting the user to write characters he want to cipher
ciphertext: bcde  // returning plaintext ciphered with the argv[1] "key" = 1

这是我的代码

int main (int argc, string argv[])
{   
    if (argc != 2)
    {
        printf("Usage: ./ceasar key\n");
        return 1;
    }
    else if (argc ==2)
        {
            int k = atoi(argv[1]);
            for (int j = 0, len = strlen(argv[1]); j < len; j++)
            {
                if (!isdigit(argv[1][j]))
                {
                    printf("Usage: ./ceasar key\n");
                    return 1;
                }

            }
            for (int j = 0, len = strlen(argv[1]); j < len; j++)
            {
                 if (isdigit(argv[1][j]))
                {
                    string s = get_string("plaintext: ");
                    printf("ciphertext: ");
                    for (int i = 0, n = strlen(s); i <= n; i++)
                    {

                        if ('@' < s[i] && s[i] < '[')
                        {
                            printf("%c", (s[i] - 'A' + k) % 26 + 'A');
                        }
                        else if('`' < s[i] && s[i] < '{')
                        {
                            printf("%c", (s[i] - 'a' + k) % 26 + 'a');
                        }
                        else
                        {
                            printf("%c", s[i]);
                        }
                        printf("\n");
                        return 0;
                    }  
                }
            }
        }
    }```


The first lines checks if argc !=2, and if argv[1][j] has a non numeric character. Once that is done it will get argv[1] and add it to each character given from the user. but it wont work correctly. 

**Any sugestions?**

Answer 1

我测试了您的代码。它只打印一个字符，这意味着最后的for loop返回得太快。该循环仅迭代一次，因为它在迭代结束时返回0。如果将return 0与printf("\n")移出for循环，则代码将起作用。

解决方案；

for (int j = 0, len = strlen(argv[1]); j < len; j++)
{
    if (isdigit(argv[1][j]))
    {
        string s = get_string("plaintext: ");
        printf("ciphertext: ");
        for (int i = 0, n = strlen(s); i <= n; i++)
        {

            if ('@' < s[i] && s[i] < '[')
            {
                printf("%c", (s[i] - 'A' + k) % 26 + 'A');
            }
            else if('`' < s[i] && s[i] < '{')
            {
                printf("%c", (s[i] - 'a' + k) % 26 + 'a');
            }
            else
            {
                printf("%c", s[i]);
            }
        }
        printf("\n");
        return 0;
    }
}

CS50 CAESAR PSET2通过命令行参数argv []

1 个答案: