Count the amount of individual characters (vowels) within a string in C - c

Question

I've just started programming in C, and I have to create a program that counts how many vowels a string has. So far I have this:
int a;
int len = strlen(text)-1
for(a=0;a==len;++a){
if(text[a]=='a'){
++vocals;}
I'm clueless on what's wrong, because it will always print 0. I understand my code as:
starting on the first char until the last one before \0,
compare them to 'a', and if they do equal, add one to the counter.
What's wrong with my code?

Answer 1

Check your tutorials or textbook on the syntax and semantics of the for loop.
It requires a continuation condition, i.e. "loop as long as this is true".
So in your code you should change to:
for(a=0; a<len; ++a)

Answer 2

change this
for(a=0;a==len;++a)
to
for(a=0;a<=len;++a)
first iteration a is not equal to len so it will never enter the loop. you want to iterate over this for loop as long as a is lesser of eual to len, second statement does precisely that.

Answer 3

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main(){
char* text = "this is a test";
int i, vocals;
int len = strlen(text);
for(i=0;i<len;++i){
if(text[i]=='a'||text[i]=='e'||text[i]=='i'||text[i]=='o'||text[i]=='u')
++vocals;
}
printf("%d\n", vocals);
}
This is a working little program.
Let's look at the main together:
First we declare a char arrays, since you omitted that part i assume your string is correctly formed.
The for loop continues till i becames equal to len. It's a good programming practice to use < or > instead of != or == to increment the strength of the code. Anyway since it's a small program you can even use != which means "Execute the for loop till i becames equal to len"
In the if statement we are just checking if the current character is equal to 'a' or 'e' or 'i' or 'o' or 'u'. As you may now guess || operator means or.
EDIT:
There are a lot of more efficient (and complex) way to do this. For example using Regular Expression. If you're interested there are a lot of good tutorial online, for example this

Answer 4

This code snippet
int a;
int len = strlen(text)-1
for(a=0;a==len;++a){
if(text[a]=='a'){
++vocals;}
does not make great sense.
For example the character 'a' is not the only vowel, the condition a == len is evaluated to true only when the string contains just one character.
You can write a separate function that counts vowels in a string.
Here is a demonstrative program.
#include <stdio.h>
#include <ctype.h>
#include <string.h>
size_t count_vowels( const char *s )
{
const char *vowels = "aeiou";
size_t count = 0;
for ( ; *s; ++s )
{
if ( strchr( vowels, tolower( ( unsigned char )*s ) ) )
{
++count;
}
}
return count;
}
int main(void)
{
char s[] = "Hello Pelput";
printf( "There are %zu vowels in the string\n\"%s\"\n",
count_vowels( s ), s );
return 0;
}
The program output is
There are 4 vowels in the string
"Hello Pelput"