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.

You can check each character at a time. So in your while loop you can have

Code:

---

if(c=='a' || c=='A')va++;

---

and so on for the other vowels.

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

Note that this

Code:

---

if(c=='a' || c=='A')va++;

---

is the same as this

Code:

---

if(tolower(c) == 'a') va++;

---

if <ctype.h> is #included. It's a bit easier to use. (There's a toupper() as well.)

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:

The trouble is that changes to your variables in CountVowelLetters() are not reflected in main(), so you don't have an accumulating count.

The easiest way to solve this is to use pointers as parameters to CountVowelLetters().

Code:

---

CountVowelLetters(c, &va, &ve, &vi, &vo, &vu);
/* ... */
void CountVowelLetters(char c, int *va, int *ve, int *vi, int *vo, int *vu){
/* ... */
if(c=='a' || c=='A') (*va)++;
/* etc */

---

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.

If you don't know about pointers, I've probably just confused you. Perhaps you should look at this pointer tutorial. http://www.cprogramming.com/tutorial/c/lesson6.html

Another solution is to put everything into main() and do away with CountVowelLetters() -- that way you don't have to use pointers. But I'm guessing you're supposed to have this function.

Quote:

---

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.

Quote:

---

Originally Posted by brewbuck

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.

---

That's true. Sigh. All those extra parentheses I could have saved, had I thought of this myself. ;)

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 );

---

or
printf("&#37;d", vowel);

and you better count number of chars stoored in the string to avoid memory overruns

proper indentation never hurts.

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!