Creating a regex for vowels in C

Question

I need to create a regex check if a word contains vowels, and after creating this check, I will cut out the vowel parts of this word, but I don’t know how to create a regex in C. I saw some codes so :

int reti;
reti = regcomp(&regex, "^a[[:alnum:]]", 0);

Could I use the same example and modify it to look like this ? Getting :

int reti;
reti = regcomp(&regex, "aeiou", 0);

Example :

Input : "algoritmo"

Output : "lgrtm"

How can I create a regex to do this job ? There is some library in C, which I can import for help in case ?

Answer 1

As you said you "need a regex", I thought to present only the regex "pure", without worrying about the language, but I would have to test it somehow, so I developed in python (I have this regex here in my "Knowledge Base" for years, it is not my own)I think it will be easy for you to pour into the.

Regex

regex = (?=[b-df-hj-np-tv-xz])(.)(?!\1)(?<!\1\1)

Explaining the parts:

(?=[b-df-hj-np-tv-xz]) Casa somente com consoantes
(.)                    A "gula" na regex, considera todas.
(?!\1) e (?<!\1\1)     Evita duplicidade no final

Python implementation:

See the code execution in repl.it.

import re
var = 'algoritmo'
r1 = r'(?=[b-df-hj-np-tv-xz])(.)(?!\1)(?<!\1\1)'
result = re.findall(r1, var)

consonants  = ''.join(str(e) for e in result)
print (consonants.split(' ')[0])
lgrtm

In reality the regex gives the result that you need but not doing as you suggest, that is, instead of identifying the vowels and removing them, it returns only the consonants.

DEMO

---

Edited Version without regex:
I’m not in time and on a very limited machine, so I developed a python version without the use of regex, I think it would be easy to convert to C, if you want to try the repl I try to help.

vogais = ['a', 'e', 'i', 'o', 'u']

string = 'Algoritmo'

result = ''

# Percorre todas as letras da string 
for x in string:
    # convert p/ minúscula e verifica se esta em vogais 
    if x.lower() not in vogais:
        # se NÃO estiver em vogais, adiciona na string resultante 
        result += x

# Imprime o resultado    
print (result)
lgrtm    

See the execution in repl.it.

Answer 2

You can do this by comparing each char of your array, in this way:

int testvogal(char p);
int main(void) {
    char regex[] = "algoritmo";
    int t = 0;
    char final[99];
    for (int i=0;i<strlen(regex);i++)
    {
        if (testvogal(regex[i]) == 1){
            final[t] = regex[i];
            t++;
        }
    }
    printf(final);
}

int testvogal(char p){
    char *vogal = "aeiou";
    for (int j=0;j<strlen(vogal);j++)
    {
        if (p == vogal[j])
            return 0;
    }
    return 1;
}

See working on Ideone.