isvowel function

Im trying to create a code that simply counts the number of vowels imputed by the user. Here's where I am.

Code:

---

#include <iostream>
 using namespace std;

 bool isVowel(char ch);

 int main()
 {

 int count=0;
 char character;
 int vowelcount=0;


 cout << "Enter a sentence"<< endl;
 cout << endl;


 while(character !='!')
 {
 cin >> character;
 

 if(isVowel(character)==true)
 {
 vowelcount++;
              cout << "Vowels are " << vowelcount << endl;
 }
 if(vowelcount !=0)
    {
        cout << endl;
            }
 }
 cout << endl;


 }

 bool isVowel(char ch)
 {

 if ('A' == ch || 'a' == ch ||
 'E' == ch || 'e' == ch ||
 'I' == ch || 'i' == ch ||
 'O' == ch || 'o' == ch ||
 'U' == ch || 'u' == ch)
 {


 }
 
 }

---

You seem to be close. What's stopping you from completing the isVowel function?

Also, you need to indent your code properly.

I'm not sure. It isn't counting them properly. If I put in a 3 letter word, it displays:

Vowels 1
Vowels 2
Vowels 3

I forgot to indent as I was working, my bad

Err... I implied that your isVowel function is incomplete: you forgot to return a value from it.

lol, thank you. I see what youre saying.

As it stands, your program shouldn't compile as isVowel(..) is supposed to return a type bool but the function doesn't have a return statement! Also, your main function should return a value as well (usually 0).

Quote:

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I forgot to indent as I was working,

---

Indenting of code as you enter/change it should become second nature. It makes reading and understanding it so much easier - especially when it comes to trying to work out why a program doesn't function as expected.

Quote:

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Originally Posted by 2kaud

As it stands, your program shouldn't compile as isVowel(..) is supposed to return a type bool but the function doesn't have a return statement!

---

It only results in undefined behaviour, hence a compiler might still compile the program. However, it probably would result in a warning, so... psfign: consider increasing the warning level of your compiler.

Quote:

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Originally Posted by 2kaud

Also, your main function should return a value as well (usually 0).

---

That's optional since a missing return statement for the global main function would be as if return 0; was present at the end.

I don't know how to do the return statement, im lost. Everytime I think im getting better, im not

Quote:

---

Originally Posted by psfign

I don't know how to do the return statement

---

Don't panic; review your learning material on functions. You will surely encounter a few examples of return statements there.

Quote:

---

I don't know how to do the return statement

---

To return a value from a function, the return statement is used. The format is

return val;

where val is the data you want to return and has the same type as that defined with the function definition. In your case isVowel(.) needs to return a value of type bool which is true if ch is a vowel and false otherwise.

This may help
http://www.learncpp.com/cpp-tutorial...-at-functions/

Quote:

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Originally Posted by psfign

I don't know how to do the return statement, im lost. Everytime I think im getting better, im not

---

In this case, as it's a bool function, all you need to do is return the result of the if statement.

Code:

---

bool isVowel(char ch)
{

    return 'A' == ch || 'a' == ch ||
          'E' == ch || 'e' == ch ||
          'I' == ch || 'i' == ch ||
          'O' == ch || 'o' == ch ||
          'U' == ch || 'u' == ch;
 }

---

ive got that part essentially, but it created an unexpected error.
23 undefined reference to `isVowel(char)'

Code:

---

#include <iostream>
 using namespace std;

 bool isVowel(char ch);
 char ch=' ';
 int main()
 {

 int count=0;
 char character;
 int vowelcount=0;


 cout << "Enter a sentence"<< endl;
 cout << endl;


 while(character !='!')
 {
 cin >> character;
 

 if(isVowel(character)==true)
 {
 vowelcount++;
    cout << "Vowels are " << vowelcount << endl;
    return 0;
 }
 if(vowelcount !=0)
 {
        cout << endl;
        return 0;
 }
 }
        bool isVowel(char ch);
 {
 if ('A' == ch || 'a' == ch ||
 'E' == ch || 'e' == ch ||
 'I' == ch || 'i' == ch ||
 'O' == ch || 'o' == ch ||
 'U' == ch || 'u' == ch)
 
 cout << endl;
 }
 }

---

You need to figure out meaningful indentation.

Code:

---

        bool isVowel(char ch);
 {
 if ('A' == ch || 'a' == ch ||
 'E' == ch || 'e' == ch ||
 'I' == ch || 'i' == ch ||
 'O' == ch || 'o' == ch ||
 'U' == ch || 'u' == ch)
 
 cout << endl;
 }

---

Lose the ;

Your function still doesn't return a value. I showed you what to do in a previous post. It certainly shouldn't cout anything.

I'd work on fixing your indenting. it will make reading your code much easier.

Ok, I tried to indent better. I also worked on the return statement, but it's not counting properly. Here's what I have so far. I cant see why it always only counts 1 vowel.

Code:

---

#include <iostream>
using namespace std;

bool isVowel(char ch);
char ch = ' ';
int main()
 {
        int count=0;
        char character;
        int vowelcount = 0;
        cout << "Enter a sentence"<< endl;
        cin >> character;cin.get(character);
        cout << endl;

        while(character !='!')
        {
                cin >> character;
 

                if(isVowel(character) == true)
                {
                        vowelcount++;
                        cout << "Vowels are " << vowelcount << endl;
                        return 0;
                }
                if(vowelcount !=0)
                        {
                                cout << endl;
                                return 0;
                        }
        }
 }
        bool isVowel(char ch)
                {
                  return 'A' == ch || 'a' == ch ||
                        'E' == ch || 'e' == ch ||
                        'I' == ch || 'i' == ch ||
                        'O' == ch || 'o' == ch ||
                        'U' == ch || 'u' == ch;
               
                        return true;
       
                }

---

Quote:

---

Originally Posted by psfign

Ok, I tried to indent better. I also worked on the return statement, but it's not counting properly. Here's what I have so far. I cant see why it always only counts 1 vowel.

---

Are you debugging your code, or just placing statements hoping the program will work?

If a program doesn't behave correctly, then it's time to learn how to debug your program. Debugging a program is part and parcel of learning how to write a program.

No one writes perfect programs the first time. Your compiler more than likely has a debugger, where you can step through the program one line at a time, watching variables and seeing how the program flows. If you did that, not only would you know what's wrong, you will learn how your program is doing what it's doing, i.e. program flow.

At the very least, you should have output the value of variables or some other statements telling you what your program is doing at certain times.

Regards,

Paul McKenzie

For example:

Code:

---

if(isVowel(character) == true)
{
    vowelcount++;
    cout << "Vowels are " << vowelcount << endl;
    return 0;
 }
 if(vowelcount !=0)
 {
    cout << endl;
    return 0;
 }

---

If you used the debugger, you would see that when the return is executed, the program ends. Then you would learn right away that the return statement did something you didn't expect, and that is to end the program. You then take out your C++ book and go to the chapter concerning "return" and the main() program. Then you see that a return from main() ends the program. All learned by starting to use the debugger.

So how do you fix it? You get rid of the return statement and place it in a more logical place in your program, e.g. where you really and truly want the program to end.

Regards,

Paul McKenzie

Quote:

---

Originally Posted by psfign

Ok, I tried to indent better. I also worked on the return statement, but it's not counting properly. Here's what I have so far. I cant see why it always only counts 1 vowel.

Code:

---

        bool isVowel(char ch)
                {
                  return 'A' == ch || 'a' == ch ||
                        'E' == ch || 'e' == ch ||
                        'I' == ch || 'i' == ch ||
                        'O' == ch || 'o' == ch ||
                        'U' == ch || 'u' == ch;
               
                        return true;
       
                }

---

---

Now you're just guessing. As Paul said, read up on what return does. I showed you exactly how to write that function.

I got it.

Code:

---

//Author: Kenneth Watkins

#include <stdlib.h>
#include <iostream>
#include <string>
using namespace std;

bool isVowel(char ch);
char ch = ' ';
int main()
 {
        int count=0;
        char character;
        int vowelcount = 0;
        cout << "Enter a sentence"<< endl;
        cin.get(character);
        cout << endl;
       
        while(character != '\n')
        {
                if(isVowel(character))
                {
                        vowelcount++;
                        cin.get(character);
                       
                }       
                else
                {
                        cin.get(character);
                       
                }
        }
                cout << "Vowels are " << vowelcount << endl;
}

        bool isVowel(char ch)
                {
                  return 'A' == ch || 'a' == ch ||
                                'E' == ch || 'e' == ch ||
                                'I' == ch || 'i' == ch ||
                                'O' == ch || 'o' == ch ||
                                'U' == ch || 'u' == ch;
                 
              }

---

Great! You can simplify the code somewhat though. You have cin.get(character) twice within the if statement when you could just do it once. As this statement is executed irrespective of the result of the if statement, you can remove the else part and put the cin.get(..) after the if. This would give

Code:

---

if (isVowel(character))
{
        vowelcount++;
}       
cin.get(character);

---

You are also doing cin.get(..) twice, once before the loop and once in the loop. If a do loop is used rather than a while then cin.get(..) is only needed once as in

Code:

---

        cout << "Enter a sentence" << endl;
       
        do {
                cin.get(character);
                if (isVowel(character)) {
                        vowelcount++;
                }       
        } while (character != '\n');

      cout << endl << "Vowels are " << vowelcount << endl

---

You can even remove the if statement altogther as isVowel returns a bool which can added to an int (true equates to 1 and false to 0) giving the final version as

Code:

---

#include <iostream>
using namespace std;

bool isVowel(char ch);

int main()
{
char        character;
int        vowelcount = 0;

        cout << "Enter a sentence"<< endl;

        do {
                vowelcount += isVowel(character = cin.get());
        } while (character != '\n');

        cout << endl << "Vowels are " << vowelcount << endl;
}

bool isVowel(char ch)
{
        return        'A' == ch || 'a' == ch ||
                        'E' == ch || 'e' == ch ||
                        'I' == ch || 'i' == ch ||
                        'O' == ch || 'o' == ch ||
                        'U' == ch || 'u' == ch;
}

---

Have a good look at the do loop and try to understand how it works.:)

Quote:

---

Originally Posted by 2kaud

If a do loop is used rather than a while then cin.get(..) is only needed once

---

I think a while loop is fine. It is a matter of conditions, e.g.,

Code:

---

while (cin.get(character) && character != '\n')
{
    if (isVowel(character))
    {
        ++vowelcount;
    }
}

---

Note that if you use cin.get() instead (i.e., no argument), then this is the version of get that returns an int, not a char, since it could return EOF (or some equivalent).

Quote:

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Originally Posted by 2kaud

You can even remove the if statement altogther as isVowel returns a bool which can added to an int (true equates to 1 and false to 0)

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You certainly could, but personally this implicit conversion makes me queasy. I would rather be explicit.