用C编写的凯撒密码程序有一个错误答案

【问题标题】：Caesar cipher program written in C has a bug用C编写的凯撒密码程序有一个错误
【发布时间】：2022-09-30 21:20:55
【问题描述】：

我为 cs50 编写了一个程序，它应该接收（整数）key（在运行时）并输出plaintext 的提示，然后根据凯撒密码函数输出明文的加密版本。当我在 cs50 IDE 中运行该程序时，它会在终端 (make caesar) 中编译，当我在运行时输入 \'key\' 时（例如./caesar 2），我会收到提示 [Plaintext: ] 和例如我输入Hello。输出将是[Ciphertext: 99102109109112]，而不是预期的[Ciphertext: JGOOQ]。

这是我的代码：


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

// Get the \'key\' value at run-time (./ceasar \'key\').
// key_value must be a digit/number).
// String argv[1] must be converted to int.
// Prompt user for plaintext.
// plaintext must be converted by casaer cipher to ciphertext.
// Print ciphertext.

// Declaring variables.
string plaintext;
string ciphertext;
int key_value;

// declaring the caesar cipher function (takes in an array of characters (aka string) and an int).
void caesar_cipher(char str[], int shift_value);

int main(int argc, string argv[])
{
    // check if there are two arguments at run-time and the second argument (argv[1]) is a digit.
    if (argc == 2 && isdigit(*argv[1]))
    {
        // convert string argv[s] to an int.
        key_value = atoi(argv[1]);
        // Prompt user for plaintext.
        plaintext = get_string(\"Plaintext: \");

        printf(\"Ciphertext: \");

        caesar_cipher(plaintext, key_value);

        // new line
        printf(\"\\n\");
        return 0;
    }
    else
    {
        printf(\"Usage: ./caesar \'key\'\\n\");
        return 1;
    }
}

// char str[] will take in \'plaintext\' and int shift_value will take in \'key\'
void caesar_cipher(char str[], int shift_value)
{
    int s = 0;
    char c = str[s];

    if (c != \'\\0\')
    {
        // iterate through every character, letter-by-letter.
        for (int i = 0, n = strlen(plaintext); i < n; i++)
        {
            // case for uppercase letters.
            if (isupper(c))
            {
                 printf(\"%i\", (((plaintext[i] - \'A\') + shift_value) % 
                        26) + \'Z\');
            }
            // case for lowercase letters.
            else if (islower(c))
            {
                printf(\"%i\", (((plaintext[i] - \'a\') + shift_value) % 26) 
                       + \'z\');
            }
            else
            {
                printf(\"%c\", c);
            }
        }
    }
}

你为什么要循环调用caesar_cipher？
else if (str[i] >= \'a\' && str[i] <= \'a\') ---> else if (str[i] >= \'a\' && str[i] <= \'z\')z！不是
@CGi03 islower() 更好...

标签： c pointers debugging cs50 caesar-cipher

【解决方案1】：

有多个问题：

您将转换后的字符输出为带有%i 的整数，而不是带有%c 的字符。
测试if (c != '\0') 是多余的：如果字符串为空，循环将立即退出。
您必须将每个字符存储到循环内的c 中才能正常运行。
您必须将 'A' 而不是 'Z' 添加到移位索引中。小写也有同样的问题。
该函数应该将加密执行到另一个字符串cyphertext，而不是打印它。

这是修改后的版本：

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

// Get the 'key' value at run-time (./ceasar 'key').
//     key value must be a number.
//     string argv[1] must be converted to int.
// Prompt user for plaintext.
// plaintext must be converted by caesar_cipher to ciphertext.
// Print ciphertext.

char *caesar_cipher(const char *str, int shift_value);

int main(int argc, string argv[]) {
    string plaintext;
    string ciphertext;
    int key_value;

    // check if there are two arguments at run-time and the second argument (argv[1]) is a digit.
    if (argc == 2 && isdigit((unsigned char)*argv[1])) {
        // convert key string argv[1] to an int.
        key_value = atoi(argv[1]);
        // prompt user for plaintext.
        plaintext = get_string("Plaintext: ");
        cyphertext = caesar_cipher(plaintext, key_value);
        printf("Ciphertext: %s\n", cyphertext);
        return 0;
    } else {
        printf("Usage: ./caesar 'key'\n");
        return 1;
    }
}

// The caesar_cipher function:
// takes in a string and a shift value,
// allocate a new string and perform the Caesar transformation in place
// returns the new string.
char *caesar_cipher(const char *str, int shift_value) {
    char *dest = strdup(str);

    // iterate through every character, letter-by-letter.
    for (int i = 0; dest[i] != '\0'; i++) {
        unsigned char c = dest[i];
        // case for uppercase letters.
        if (isupper(c)) {
            c = 'A' + (c - 'A' + shift_value) % 26;
        } else
        if (islower(c)) {
            c = 'a' + (c - 'a' + shift_value) % 26;
        }
        dest[i] = c;
    }
    return dest;
}

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