How can we get vowel substring from a given string

Question

I want to get all vowel substrings from the given string. Given string is 'auiouxaeibaou', get substrings from the given string like [auiou,aei,aou].

Here I tried something like this, but not getting the exact output.

bool isVowel(char c) {
    return (c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u');
}

void substr(char str[], int low, int high)
{
    printf("%.*s \n\n ", high-low+1, (str+low));
}

int main(int argc, const char *argv[]) {

    char str[] = "aeixae";

    int length = strlen(str);

    int start_index = 0, end_index = 0;

    for (int x=0; x<length; x++) {
        char c  = str[x];
        if (isVowel(c) == false) {
            end_index = x;
            substr(str, start_index, end_index - 1 );
            start_index = end_index + 1;
        }

    }
    return 0;
}

Answer 1

For fun, here is a simple way to do it without substrings, just print vowels as they come and add newlines the first time you hit a consonant.

#include <stdio.h>
#include <string.h>

int isVowel(char c){
    if(c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u'){
        return 1;
    }
    else{
        return -1;
    }
}

int main()
{       
   char my_str[] = "auiouxxxxxxaeibaou";

   int todo_newline = 1; // prevent consecutive newlines

   for(int i = 0; my_str[i] != '\0'; i++){
       if(isVowel(my_str[i]) == 1){
           printf("%c", my_str[i]);
           todo_newline = 1; 
       }
       else{
           if(todo_newline == 1){ // print one newline
               printf("%c", '\n');
               todo_newline = -1; // don't print consecutive newlines
           }
       }
   }
}

Answer 2

Alternative approach:

strspn(), strcspn() are the best tools here. @Gem Taylor

#include <stdio.h>
#include <string.h>

int main(void) {
  char str[] = "aeixae";
  const char *s = str;
  while (*(s += strcspn(s, "aeiou")) != '\0') { // skip letters that are not aeiou
    size_t len = strspn(s, "aeiou");            // find length made up of aeiou
    printf("<%.*s>\n", (int) len, s);           // print sub string
    s += len;
  }
}

Output

<aei>
<ae>

Answer 3

your attempt was close. I just added the includes and made sure that the last part of the string also was printed. See the termination of the for loop and the if where you check for the vowel.

#include <stdio.h>
#include <string.h>
#include <stdbool.h>

bool isVowel(char c) {
    return (c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u');
}

void substr(char str[], int low, int high)
{
    printf("%.*s \n\n", high-low+1, (str+low));
}

int main(int argc, const char *argv[]) {

    char str[] = "aeixae";

    int length = strlen(str);

    int start_index = 0, end_index = 0;

    for (int x=0; x<=length; x++) {
        if (x == length || !isVowel(str[x])) {
            end_index = x;
            substr(str, start_index, end_index - 1 );
            start_index = end_index + 1;
        }

    }
    return 0;
}

and this is the output:

gcc main.c && ./a.out 
aei 

ae 

Answer 4

but not getting the exact output.

why you don't have the expected result :

• you miss the last sequence of vowel, to have it just replace for (int x=0; x<length; x++) by for (int x=0; x<=length; x++) because the null character is not a vowel (that does not produce illegal access)
• when you have several consecutive non vowel you call anyway substr, to avoid that you need to memorize if had or not previously a vowel

The modifications gives that (I changed the input string) :

bool isVowel(char c) {
    return (c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u');
}

void substr(char str[], int low, int high)
{
    printf("%.*s \n\n ", high-low+1, (str+low));
}

int main(int argc, const char *argv[]) {
  char str[] = "aeixaewwii";

  int length = strlen(str);

  int start_index = 0, end_index = 0;
  bool wasVowel = false;

  for (int x=0; x<=length; x++) {
    char c  = str[x];
    if (isVowel(c) == false){
      end_index = x;
      if (wasVowel)
        substr(str, start_index, end_index - 1 );
      start_index = end_index + 1;
      wasVowel = false;
    }
    else
      wasVowel = true;
  }

  return 0;
}

BTW : 'y' is a vowel for me, you missed it in isVowel()

Answer 5

1. Nice program. Remember to use size_t as the proper type returned by strlen().
2. Well, if you want input to be auiouxaeibaou, you need to insert it at char str[] = "aeixae";
3. Great substr function!
4. You need to remember about the last substr from the string - when x reaches lenth, there is still one substring

#include <string.h>
#include <stdio.h>
#include <stdbool.h>


bool isVowel(char c) {
    return (c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u');
}

void substr(char str[], int low, int high)
{
    printf("%.*s\n", high-low+1, (str+low));
}

int main(int argc, const char *argv[]) {

    // ch-ch-ch-changes   
    char str[] = "auiouxaeibaou";

    int length = strlen(str);

    int start_index = 0, end_index = 0;

    for (int x = 0; x < length; x++) {
        char c  = str[x];
        if (isVowel(c) == false) {
            end_index = x;
            substr(str, start_index, end_index - 1 );
            start_index = end_index + 1;
        }
    }

    // ch-ch-ch-changes
    end_index = length;
    substr(str, start_index, end_index - 1 );

    return 0;
}