CODEWITHSUNDEEP
cmacros
CaptainDaVinci
CaptainDaVinci

Reputation: 1065

How can I define a macro that tests if a given character is a vowel?

The following, code snippet does not give compilation error, but it does not give the expected output either, though this could be done in simple if-else way but I wanted to do it using macros. Here c is a character variable.

#define VOWELS 'a' || 'e' || 'i' || 'o' || 'u' || 'A' || 'E' || 'I' || 'O' || '
if (c == VOWELS) {
   printf("vowel = %c\n", c);
}

Upvotes: 4

Views: 501

Answers (2)

Rishikesh Raje
Rishikesh Raje

Reputation: 8614

This will expand to 

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

Which will check for c==a and then logical OR it with e which has a non zero value. So the result will always be TRUE.

What you want is 

 #define VOWELCHECK(c) ((c)=='a') || ((c)=='e') || ((c)=='i') || \
                       ((c)=='o') || ((c)=='u') || ((c)=='A') || \
                       ((c)=='E') || ((c)=='I') || ((c)=='O') || ((c)=='U')))


// In the program
if (VOWELCHECK(c))
{
   printf("vowel = %c\n", c);
}

Upvotes: 3

Govind Parmar
Govind Parmar

Reputation: 21562

That's because everything but the leftmost value in the VOWELS macro is not being tested against c. What the macro expands to is:

c == 'a' || 'e' || ...

So basically, since a non-zero expression (i.e., the numeric value of the character 'e') is being tested for, that always evaluates to 1.

What the macro should be is:

#define VOWEL(c) ((c) == 'a') || ((c) == 'e') || ((c) == 'i') || ((c) == 'o') || ((c) == 'u') || ((c) == 'A') || ((c) == 'E') || ((c) == 'I') || ((c) == 'O') || ((c) == 'U')

And then, you would simply use:

if(VOWEL(c))
{
    ...
}

Upvotes: 7

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