Counting vowels in string. "Comparison pointer-integer".


 
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# 1  
Question Counting vowels in string. "Comparison pointer-integer".

I'm trying to write a programme which scans strings to find how many vowels they contain. I get an error saying that I'm trying to compare a pointer and an integer in
Code:
if(*v == scanme[i]){

. How can I overcome this ? Also, the programme seems to scan only the first word of a string e.g.: if I type "abc de" it will only scan "abc". (I know this because for some reason it didn't produce the pointer-integer error before).

Thank you in advance.

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

int main ()
{
        char scanme[80];
        char *v[80] = {"a", "e", "i", "o", "u"};
        int nv = 0;
        int i;
        printf ("Type a string to scan.\n");
        scanf ("%s", scanme);
        printf ("Looking for vowels in \"%s\"...\n", scanme);


        for(i = 0; i < strlen(scanme); i++){
                if(*v == scanme[i]){
                        nv++;
                }
        }
        printf ("There are %d vowels.\n", nv);

        return 0;
}

# 2  
just a guess..

at line where it says
Code:
      if(*v == scanme[i]){

i don't think you are comparing each element of v. v is a pointer to an array. You may need to use something like v[counter], maybe in a nested loop.

hth

Last edited by laiko; 12-06-2011 at 01:42 AM..
 

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