problem with letters

I have this code but it seems no matter what letter I choose it goes to if and not else:

Code:

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void consonant()
{
        do
        {
                printf("Pick a letter: ");
                        letter=getch();
                if (letter == 'a', 'e', 'i', 'o', 'u')
                        printf("You can't pick a vowel choose again!!\n");
                else
                        printf("\nYou choosed %c.\n\n", letter);
                        valid = 1;
        }
        while(valid == 0);
       
        endturn = 0;
        winner = 0;
}

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Thanks for the help

Quote:

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if (letter == 'a', 'e', 'i', 'o', 'u')

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I don't think that's legal. You either have to say:

Code:

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if (letter == 'a' || letter == 'e' || ...)

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or you can write a macro or a function to do that for you:

Code:

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#define is_vowel(x) ( x == 'a' || x == 'e' || x == 'i' ...)

if (is_vowel(letter)) ...

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Quote:

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Originally posted by Salem
> I don't think that's legal. You either have to say:
Oh its quite legal - its just pretty useless - and almost certainly not what was wanted.
It basically degenerates into if ( 'u' != 0 )

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Why exactly would it degenerate into that? I don't see why thatis the case, although I can see that the initial statement was wrong.

>Why exactly would it degenerate into that?

From C99:

Quote:

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The left operand of a comma operator is evaluated as a void expression; there is a sequence point after its evaluation. Then the right operand is evaluated; the result has its type and value.

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So, first letter == 'a' is first evaluated as a void expression, then 'e' is evaluated as a void expression, then 'i' is evaluated as a void expression, then 'o' is evaluated as a void expression, then 'u' is evaluated. This is the condition in the if statement, whether 'u' is nonzero. (And that would be a pretty safe bet!)