Find Word with Vowels in order C - c

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!

Answer 2

but its flawed somewhere...
Could there be an error here?
if ( *vowel == 'a' &&
(*vowel + 1) == 'e' &&
(*vowel + 2) == 'i' &&
(*vowel + 3) == 'o' &&
(*vowel + 4) == 'e' )
// ^^^ 'u'?
There may also be other errors. I haven't checked all your code.

Answer 3

Here is a big flaw:
char *v = malloc(sizeof(char *));
This only allocates four or eight bytes (depending on if you are on a 32 or 64 bit platform). I'm guessing you want a little more than that.
PS. In the future, you should probably try to be more specific instead of just saying that "it's flawed".

Answer 4

Why is all_vowels() allocating memory? And, even more interesting, why doesn't it free() it?
I'm pretty sure all_vowels() doesn't have to allocate any memory, and can be a bit simpler than what you have.
Also, you can´t use feof() before trying to read from the file. Remove that, just loop until fgets() returns NULL.
I would probably write a helper function int is_vowel(char c); to make the code clearer, and then attack the problem like so in all_vowels():
vowels = "aeiou"
for each character x in the string to check:
if is_vowel(x):
if vowels starts with x:
let vowels = vowels[1:]
else
return false
return true if vowels is empty

Answer 5

Okay.. So I finally got the right output... Any hints or tips for greater efficiency would be much appreciated.
int all_vowels( char *s )
{
const unsigned char *p = (const unsigned char *)s;
char v[5];
int i = 0;
if(*p == '\0') return -1;
while( *p != '\0' )
{
if( (*p == 'a' || *p =='e' || *p =='i'|| *p =='o' || *p =='u') && ( i < 5 ) )
{
v[i] = *p;
i++;
}
p++;
}
if ( ( v[0] == 'a' && v[1] == 'e' && v[2] == 'i' && v[3] == 'o' && v[4] == 'u' ) && (strlen(v) == 5 ))
{
return 1;
}
return -1;
}
int main (int argc, char *argv[])
{
FILE *file;
char line[30];
if (( file = fopen("/usr/share/dict/words", "r") ) == NULL)
{
fprintf(stderr, "cannot open %s\n", "/usr/share/dict/words");
exit(1);
}
while ( fgets(line, sizeof(line), file) )
{
if ( all_vowels(line) == 1 )
{
printf("%s\n", line);
}
}
fclose(file);
return 0;
}

Answer 6

little unsure regarding parseing your file yet i function(s) below check that a character is a vowel and tests whether the next vowel is greater then the current vowel.
#include <stdio.h>
// for readability not advocating the
// usage of #define booleans etc
#define TRUE 1
#define FALSE 0
int isVowel (char c)
{
switch (c)
{
case 'a': return TRUE;
case 'e': return TRUE;
case 'i': return TRUE;
case 'o': return TRUE;
case 'u': return TRUE;
case 'A': return TRUE;
case 'E': return TRUE;
case 'I': return TRUE;
case 'O': return TRUE;
case 'U': return TRUE;
}
return FALSE;
}
int hasOrderedVowels (char *str)
{
char c1, c2;
c1 = *str;
c2 = *(++str);
// ignore words beginning in vowels other then 'a' or 'A'
if (isVowel(c1) && !(c1 == 'a' || c1 == 'A')) return FALSE;
do {
// ignore case of `c1`
if (c1 >= 'a')
c1 -= 32;
// ignore case of `c2`
if (c2 >= 'a')
c2 -= 32;
// compare vowels and increment
// pointers as appropriate
if (isVowel(c1) && isVowel(c2))
{
// if we have found a vowel less then or equal to current
// then they are not in order/more then one, if we have found
// a 'U' and there are more vowels then this would be a duplicate
if (c2 <= c1 || c1 == 'U')
return FALSE;
c1 = c2;
}
else if (isVowel(c2)) // found first vowel so assign to c1
{
if (!(c1 == 'a' || c1 == 'A'))
{
return FALSE;
}
c1 = c2;
}
else if (!isVowel(c1))
{
c1 = *(str += 2); // skip over c2
}
c2 = *(++str);
}
while (c2 != '\0');
return (c1 == 'U');
}
int main ()
{
char *str[] = {"aeiou", "facecious", "chimpanze", "baboon"};
int i = 0;
for (; i<5; i++)
{
printf ("%s: %i\n", str[i], hasOrderedVowels(str[i]));
}
return 0;
}
demo