...still need some help!

OK, I think that I am getting closer to completing this simple program which identifies whether or not a letter entered by the user is a vowel or not by using a function. The program successfully returns whether or not it is a vowel however it prints the lines in a disorderly fashion and it also shows some smiley face after the line "the letter you have entered is: " If you guys could help a NOOB clean this up it would be much appreciated. I am amazed how confused this stuff is making me

Code:

#include <iostream>
using namespace std;

char isVowel(char ch);

int main()
{
	char letter;

	cout << "Please enter a letter from the alphabet: ";
	cin >> letter;
	cout << "the letter you have entered is: "<<isVowel(letter)<<""<<endl;
	
	return 0;
}
char isVowel(char ch)
{
	if (ch == 'A' || ch == 'E' || ch == 'I' || ch == 'O' || ch== 'U'||
		ch == 'a' || ch == 'e' || ch == 'i' || ch == 'o' || ch== 'u')
		cout << "a Vowel" <<endl;
	else  
		cout << "not a vowel" <<endl;
	return 1;    
}

you should use the toupper() or tolower() function.. located in the <cctype> library.. to limit your user entries to either upper or lower case.. instead of listing all vowels in both upper and lower case...

Code:

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

                               cout << "a Vowel" <<endl;

could be:

Code:

toupper(ch);     //turn 'ch' to upper case

if (ch ==  'A' || 'E' || 'I' || 'O' || 'U')
		
		cout << "a Vowel" <<endl;

save yourself a little bit of code

Code:

int v( int c )
{
    int d = toupper( c );
#define C case
    switch( d ){C 'A':C 'E':C 'I':C 'O':C 'U':return 1;}return 0;
}

[edit]

Code:

int v( int c )
{
        int d=toupper(c);
        return (d=='A')|(d=='E')|(d=='I')|(d=='O')|(d=='U');
}

Parenthesis to stave off warnings...
[/edit]
Quzah.

I've got a lot of research and learning to do, the answer you gave me quzah I understand absolutely none of...except maybe return 0...lol

Quzah was just messing with you its called a macro to replace something in your code with what you defined eg in quzah's case he defined C as case so every where there is a C in your code the compiler replaces it with case

Originally Posted by The Brain

could be:

Code:

toupper(ch);     //turn 'ch' to upper case

if (ch ==  'A' || 'E' || 'I' || 'O' || 'U')
		
		cout << "a Vowel" <<endl;

save yourself a little bit of code

Actually, no, it couldn't. Try it.

Quzah.

oi...

Code:

if (toupper(ch) ==  'A' || 'E' || 'I' || 'O' || 'U')
		
		cout << "a Vowel" <<endl;

i guess toupper() or tolower() do not modify the variable.. it only returns the upper/lower case version of it's argument.

learn something new everyday.

You're still doing it wrong. Look at everything to the right of the first ||. That's all wrong. Put that as-is in a function, and pass it 'D' and watch it say a Vowel.

Quzah.

Originally Posted by quzah

Code:

        return (d=='A')|(d=='E')|(d=='I')|(d=='O')|(d=='U');

Quzah.

I think you mean this...

Code:

if (toupper(ch) ==  'A' | 'E' | 'I' | 'O' | 'U')

I am not familiar with the | operator.. but why use this as opposed to the binary || OR operator..

I think it would work with using the || operators.. i know i have used similar logic before in my programs and it works fine.



[edit]: I think | appears to be an "exclusive OR" operator

Run this:

Code:

#include <iostream>
int main()
{
    int d = 'D';
    if( d == 'E' || 'F' )
        std::cout << "See now? << std::endl;
    return 0;#include <iostream>
int main()
{
        int d = 'D';
        if( d == 'E' || 'F' )
                std::cout << "See now?" << std::endl;
        return 0;
}

Go run mine and you'll see that mine works as intended while yours doesn't. You probably won't understand why, but hopefully the above will make it 'click'. If not, I'll tell you next post.

Quzah.

Code:

#include <iostream>

int main()
{
        int d = 'D';
        if( d == 'E' || d == 'F' )
                std::cout << "See now?" << std::endl;
        return 0;
}

man... how did I miss this..>?!?!?



must be getting late zzZZzZ



How about this..??

Code:

if (toupper(ch) == ('A' || 'E' || 'I' || 'O' || 'U'))
		
		cout << "a Vowel" <<endl;

Thanks for making me use my brain btw

How about

Code:

char isVowel(char ch) {
  return strchr( "AEIOU", toupper(ch) ) != NULL;
}

Spooky - A deja-vue moment.

Geez, somebody just tell him already

Code:

if (toupper(ch) == ('A' || 'E' || 'I' || 'O' || 'U'))

You can't use || like that. Salem's answer works. Yours could go like this:

Code:

char upCh = toupper(ch);
if (upCh == 'A' || upCh == 'E' || upCh == 'I' || upCh == 'O' || upCh == 'U'))

For strgglingstuden:

The problem is that you are returning 1 in a function that returns a char. Then you are printing out the char with the ASCII value of 1, which I guess is a smiley face. Since your function already outputs whether the character is a vowel or not, you don't need to return anything, and you certainly don't need to cout the return value of the function.

If you want, you could remove the cout statements from your function and return a bool instead of a char. Then, if the letter is a vowel (your if statement), return true, and otherwise return false. Then in your main function have an if (isVowel(letter)) that outputs the correct statement at the end of your sentence. This is the preferred method of using function. The isVowel function would have one purpose, to return true or false depending on whether the character is a vowel or not.