Thread: Counting Vowels within a string.

Hi, I have created a program counting vowels in a string

Vowels (A, E, I, O, U, a, e, i, o, u).

I don't know how to get it to show something like this:

Input: This is a string.

Output: The number of Vowels is: 4

Code:

#include <stdio.h>
#define SIZE 50

int main (void)
{

int va = 0;              /*Vowel counters*/
int ve = 0;
int vi = 0;
int vo = 0;
int vu = 0;

char c;                   /*Character input to be placed into array*/
char s[SIZE];         /*Character Array*/
int i=0;

puts("Enter a Line of Text:");

                      
while ((c = getchar()) != '\n'){
s[i++] = c;
}
                        
                       
    

puts ("\n");
puts(s);
printf("\n");
printf("%d", a);
printf("\n");

return 0;
}

Any help would be greatly appreciated.

This is perfect, thank you very much. That is exactly what I needed.

Code:

#include <stdio.h>
#define SIZE 50


void CountVowelLetters(char, int, int, int, int, int);
void CountVowelLetters1(int, int, int, int, int);

int main (void)
{
int vowels;
int va = 0;
int ve = 0;
int vi = 0;
int vo = 0;
int vu = 0;
char c;
char s[SIZE];
int i=0;

puts("Enter a Line of Text:");

while ((c = getchar()) != '\n'){
s[i++] = c;
CountVowelLetters(c, va, ve, vi, vo, vu);
}

puts("The number of Vowels A,E,I,O,U,a,e,i,o,u is:");

return 0;
}

void CountVowelLetters(char c, int va, int ve, int vi, int vo, int vu){
int vowel;

if(c=='a' || c=='A')va++;
if(c=='e' || c=='E')ve++;
if(c=='i' || c=='I')vi++;
if(c=='o' || c=='O')vo++;
if(c=='u' || c=='U')vu++;

vowel = (va + ve + vi + vo + vu);
printf("%d", vowel);

}

Have it all set up, but I don't understand how to do this:

Input:
This is ok

Output:
The number of Vowels A,E,I,O,U,a,e,i,o,u is: 3

Instead i have
Output:
This is ok
0010010010The number of Vowels A,E,I,O,U,a,e,i,o,u is:

Originally Posted by dwks

The parentheses around (*va) are necessary because ++ has a higher precedence than va. For example, just plain

Code:

*va++;

is like

Code:

*(va ++);

which does not do what you want.

Or, you could just write ++*va, which is unambiguous because both unary operators are binding on the same side, therefore precedence doesn't even enter the equation.

I thnk i understand what you guys are saying, if i put it into a pointer i can call it back later in the program. I would like to thank you guys for such quick responses. This is a much better understanding then my text.

Ok this is what I think it is supposed to be but not working still.

Code:

#include <stdio.h>
#define SIZE 50


void CountVowelLetters(char, int *, int *, int *, int *, int *, int *);

int main (void)
{
int *vPtr; 
int *aPtr;
int *ePtr; 
int *iPtr; 
int *oPtr;
int *uPtr;

int vowel = 0;
int va = 0;
int ve = 0;
int vi = 0;
int vo = 0;
int vu = 0;
 
char c;
char s[SIZE];
int i=0;

aPtr = &va;
ePtr = &ve;
iPtr= &vi;
oPtr = &vo;
uPtr = &vu;
vPtr = &vowel;

puts("Enter a Line of Text:");

while ((c = getchar()) != '\n'){
s[i++] = c;

CountVowelLetters(c, &va, &ve, &vi, &vo, &vu, &vowel);

}

puts("The number of Vowels A,E,I,O,U,a,e,i,o,u is:");
printf("&#37;d", vPtr );


return 0;  
}

void CountVowelLetters(char c, int *va, int *ve, int *vi, int *vo, int *vu, int *vowel){

if(c=='a' || c=='A')(*va)++;
if(c=='e' || c=='E')(*ve)++;
if(c=='i' || c=='I')(*vi)++;
if(c=='o' || c=='O')(*vo)++;
if(c=='u' || c=='U')(*vu)++;
 
*vowel = (*va + *ve + *vi + *vo + *vu);
}

I gives me a -4197080 number every time.

I am so close...........please inform what the heck I have done wrong here.

Code:

printf("%d", *vPtr );

I guess that would make sense.

i took out

Code:

int *vPtr; 
int *aPtr;
int *ePtr; 
int *iPtr; 
int *oPtr;
int *uPtr;

aPtr = &va;
ePtr = &ve;
iPtr= &vi;
oPtr = &vo;
uPtr = &vu;
vPtr = &vowel;

Thanks guys, you are making this very easy to understand. It works properly!