c - PSET 2 : Vigenere Cipher partially working?

Question

我创建了以下代码作为对 CS50x PSET2: Vigenere 的回答，它在某种程度上有效，但是当运行 check50 时，我得到了下面列出的一些错误:

:) vigenere.c exists.
:) vigenere.c compiles.
:) encrypts "a" as "a" using "a" as keyword
:( encrypts "barfoo" as "caqgon" using "baz" as keyword - output not valid ASCII text 
:( encrypts "BaRFoo" as "CaQGon" using "BaZ" as keyword - output not valid ASCII text 
:) encrypts "BARFOO" as "CAQGON" using "BAZ" as keyword
:( encrypts "world!$?" as "xoqmd!$?" using "baz" as keyword- output not valid ASCII text 
:( encrypts "hello, world!" as "iekmo, vprke!" using "baz" as keyword- output not valid ASCII text 
:) handles lack of argv[1]
:) handles argc > 2
:( rejects "Hax0r2" as keyword - timed out while waiting for program to exit 

似乎发生的情况是 key 包含高值(即 z/Z)，它导致代码跳到下一行并错过看似随机序列的内容。例如。在字符串的第一个单词中，它错过了第 3 个字符，然后是第二个单词，它错过了第 3 个和第 4 个字符，然后是第 3 个单词第 1 个。我只是不明白发生了什么。

我使用 printf 来确保设置和传递给函数的所有变量在运行时都是正确的。函数本身返回正确的响应(Hax0r2 的验证除外)。我尝试通过将结果与在线 vigenere 密码工具进行比较来进行调试。

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

int Validate1(int argc);
int Validate2(string argv);
void Cypher(string x);
void KeyCalc(string argv);

string MESSAGE;
int LENGTH;
int *KEY;
int COUNTER = 0;

int main(int argc, string argv[])
{
    //Check if right amount of arguments are supplied
    int Val1 = Validate1(argc);

    if (Val1 == 0)
    {
        //Check if argument is a string of chars
        int Val2 = Validate2(argv[1]);

        if (Val2 == 0)
        {
            //get the string length
            LENGTH = strlen(argv[1]);

            //Dynamically update KEY array length
            KEY = (int *)malloc(LENGTH * sizeof(*KEY));
            if (KEY == NULL)
            {
                fprintf(stderr, "malloc failed\n");   
            }

            //calculate the key
            KeyCalc(argv[1]);

            //get the message from the user to be encrypted
            MESSAGE = get_string("plaintext: ");
            printf("ciphertext: ");

            //encrypt message from user
            Cypher(argv[1]);
            free(KEY);
            return 0;
        }
        else
        {
            //validation failed
            printf("Usage: ./vigenere keyword\n");
            return 1;
        }
    }
    else
    {
        //validation failed
        printf("Usage: ./vigenere keyword\n");
        return 1;
    }
}

//Validate the number of arguments supplied
int Validate1(int argc)
{
    if (argc != 2)
    {
        return 1;
    }
    else
    {
        return 0;   
    }
}

//Validate the argument is a string
int Validate2(string argv)
{
    int k = 0;

    //loop through all characters in argument line string and check if alphabetic 
    for (int i = 0; i < LENGTH; i++)
    { 
        if isalpha(argv[i])
        {
            //Do Nothing
        }
        else
        {
            k++; 
        }
    }

    //k counts the number of non-alphabetic characters, so if > 0 then invalid input
    if (k > 0)
    {
        return 1;
    }
    else
    {
        return 0;    
    }
}


void Cypher(string x)
{
    //identify the length of the message to be coded
    int Mlength = strlen(MESSAGE);

    //identify the length of the key
    int Slen = strlen(x);

    //cycle through all characters in message supplied by user
    for (int i = 0; i < Mlength; i++)
    {
        // loop through key
        if (COUNTER > Slen - 1)
        {
            COUNTER = 0;
        }
        //check if the character is alphabetic
        if (isalpha(MESSAGE[i]))
        {
            //convert the character to ASCII int value
            char l = MESSAGE[i];

            //add key value to message value and wrap around ascii mapping
            if (isupper(MESSAGE[i]))
            {
                l = l + KEY[COUNTER];
                if (l > 'Z')
                {
                    l = l - 26;    
                }
            }
            else
            {
                l = l + KEY[COUNTER];
                if (l > 'z')
                {
                    l = l - 26;    
                }    
            }

            //convert value back into character and store in array
            MESSAGE[i] = (char) l;
            // print character 
            printf("%c", MESSAGE[i]);
            COUNTER++;
        }
        else
        {
            //character is 'numeric' or 'symbol' or 'space' just display it
            printf("%c", MESSAGE[i]);
        }
    }
    printf("\n");
}

void KeyCalc(string argv)
{
    //convert key entry to values A/a = 0 to Z/z = 26
    for (int i = 0; i < LENGTH; i++)
    {
        char k = argv[i];
        if (islower(argv[i]))
        {
            KEY[i] = k - 'a'; 
        }
        else
        {
            KEY[i] = k - 'A'; 
        }      
    }    
}

• 使用“baz”作为关键字将“barfoo”加密为“caqgon”
• 使用“BaZ”作为关键字将“BaRFoo”加密为“CaQGon”
• 加密“world!$?”作为“xoqmd!$？”使用“baz”作为关键字
• 加密“你好，世界!”作为“iekmo，vprke!”使用“baz”作为关键字
• 拒绝将“Hax0r2”作为关键字

Answer 1

来自 caesar pset 的规范:

...Caesar’s algorithm (i.e., cipher) encrypts messages by “rotating” each letter by k positions. More formally, if p is some plaintext (i.e., an unencrypted message), p_i is the ith character in p, and k is a secret key (i.e., a non-negative integer), then each letter, c_i, in the ciphertext, c, is computed as

c_i = (p_i + k) % 26

这个算法(在任何一种“情况下”)都不会这样做:

 l = l + KEY[COUNTER];
                if (l > 'Z')
                {
                    l = l - 26;    
                }

This walkthrough从 9:30 开始是关于如何实现“转变”的一个很好的入门。

此代码中问题的最直接原因是此 l = l + KEY[COUNTER]; 可以产生 ascii range 之外的结果.在 CS50 实现中，char 默认为 signed 字符。因此，例如，'r' + 'z'(如“barfoo”中用“baz”加密)将产生 -117。