Question

So for a lab at uni… Ive been challenged to find all words in the usr/share/dict/linux.words
file using fopen, fgets etc with every vowel only once, in order.

i.e. facetious

So far I have the following code… but its flawed somewhere…

int all_vowels( char *s )
{ 
    const unsigned char *p = (const unsigned char *)s;

    char *v = malloc(sizeof(char *));
    char *vowel = v;

if(*p == '\0') return -1;

while( *p != '\0' )
{    
        if( *p == 'a' || *p =='e' || *p =='i'|| *p =='o' || *p =='u' )
        {
            *v = *p;
            v++;
        }
        p++;
    }


    if ( *vowel == 'a' && (*vowel + 1) == 'e' && (*vowel + 2) == 'i' && (*vowel + 3) ==     'o' && (*vowel + 4) == 'u' ) 
    { 
        return 1; 
    }

    return -1;
}

int main (int argc, char *argv[])
{    
    FILE *file;
    char line[BUFSIZ];

    if (( file = fopen("/usr/share/dict/words", "r") ) == NULL) 
    {
        fprintf(stderr, "cannot open %s\n", "/usr/share/dict/words");
        exit(1);
    } 

    while ( !feof(file) )
    {
        fgets(line, sizeof(line), file);
        if ( all_vowels(line) == 1 )
        {
            printf("%s\n", line);
        }
    }
    fclose(file);
    return 0;

}

Any tips would be great!!!

Im really confused at the moment…

Answer 1

You are accessing v as if it were pointing to a location holding a number of characters, when indeed you only reserve space for one single char * (usually 4 byte on a 32 bit machine and 8 byte on a 64 bit machine):

char *v = malloc(sizeof(char *));

That might or might not be enough for what you are trying to store in it; in your case, the number of vowels, in any given word.

Whenever possible, you should avoid dynamic allocations; in your case, you don’t need them, you can declare an array of fixed size instead of a char*:

char v[5];

In addition to that, you have to check if you have read 5 vowels already, so that you don’t exceed the array size; if, after 5 vowels, you encounter another one, you can stop the check anyway; the currently encountered one has to be a duplicate vowel, the word will therefore not qualify.

The way you address characters is also a problem. Check again what * does: it dereferences the expression immediately to the right. In your case, it will always dereference v, then add something to it (which is also legal, since the result of dereferencing is a char). So if the first character where v points to is an a, the second an e, then *v will yield 'a', (*v + 1) will yield 'b', (*v +2) will yield 'c' and so on – you see, the result is an addition to the letter a by the given number; it doesn’t matter at all what comes after the first character because you never access the value there. To achieve what you want with pointer arithmetic, you’d have to use parenthesis: *(v+1) – i.e., add 1 to the pointer v, then dereference it. This would yield the second character in the c string starting at v, i.e. 'e'.
Note that with v declared as above you can simply write v[0], v[1], v[2] and so on to address each character.

Aside from that, check the last comparison in your if condition, you had an ‘e’ instead of an ‘u’ there.

By the way, as a side note, and something to think about: There is a solution to your problem which does not require the v/vowel variables at all… only a single integer variable!