CODEWITHSUNDEEP
arraysc
Pool_one
Pool_one

Reputation: 1

Counting the number of vowels in a string in C

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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);
        
    }
}

Upvotes: 0

Views: 232

Answers (2)

0___________
0___________

Reputation: 67713

  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));
}

Upvotes: 0

Lundin
Lundin

Reputation: 214265

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 */.

Upvotes: 1

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