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I'm solving the decryption problem in C language

There's a problem.

There's a process of counting the vowels in the string,

code not reading the number of vowels properly in that 'countingmeasure'

I was so curious that I debugged it,

count doesn't work at'o'.

I'm really curious why this is happening

#include <stdio.h>
int main(void)
{

    char original[15] = { 't','f','l','e','k','v','i','d','v','r','j','l','i','v',NULL };


    printf("암호화된 문자열 : %s\n", original);
    printf("원본 가능 문자열  \n");
    printf("\n");



    for (int j = 0; j < 26; j  )//모음이 7개일때만 출력을 어떻게 시킬까?
    {
        char change[14] = { 0 };
        int counter=0;
        char a;
        

        for (int i = 0; i < 14; i  )
        {   
            a = original[i]   j;

            if (a > 122)
            {
                original[i] -= 26 ;
            }

            if (a == 'a' || a == 'e' || a == 'i' || a == 'o' || a == 'u')
            {
                counter  ;
            }           


            
            printf("%c", original[i]   j);
        }
        printf(" %d\n",counter);
        
    }
}

CodePudding user response：

a = original[i] j; doesn't make any sense, since a is a char and the result might not fit inside it. Specifically, "character value 26" might be larger than 127. Is char signed or unsigned by default?

Furthermore, arithmetic on any symbols except '0' to '9' isn't well-defined and they are not guaranteed to be allocated adjacently. Also please refrain from using hard-coded "magic numbers" in source code. Instead of 122 you should use 'z' etc.

There are several ways you can fix the program.

• The quick & dirty solution is to do unsigned char a on the existing program, if you are content with "it just works, but I don't even know what I'm doing".
• A better solution is to declare a string of vowels and then for every character in the input string, do a strchr() search in the vowel string for a match. (Correct but naive and slow, good enough beginner solution.)
• A professional solution would be to create a look-up table of 128 booleans like
  const bool LOOKUP [128] = { ['A'] = true, ['a'] = true, ['E'] = true, ... }; Then check if(item[i] == LOOKUP[ item[i] ]) /* then vowel */.

CodePudding user response：

1. Use functions.
2. NULL is a pointer not zero or null terminating character
3. Use string literals.
4. use standard functions to change the case (tolower, toupper)

char *mystrchr(const char *str, const char lt, int ignoreCase)
{
    while(*str) 
    {
        if(ignoreCase) 
            if(tolower((unsigned char)*str) == tolower((unsigned char)lt)) return (char *)str;
        else 
            if(*str == lt) return (char *)str;
        str  ;
    }
    return NULL;
}

size_t count(const char *haystack, const char *needle, int ignoreCase)
{
    size_t count = 0;
    while(*haystack)
    {
        if(mystrchr(needle, *haystack, ignoreCase)) count  ;
        haystack  ;
    }
    return count;
}

int main(void)
{
    char *str = "tflekvidvrjliv";

    printf("%zu\n", count(str, "aeiou", 1));
}