Weird macro definition issue

Question

I want to define a macro at compile time based on the value of another macro. However this code is not executing as expected:

#include <stdio.h> #include <stdlib.h> #include <string.h> #define SIXTEEN 16 #define TWO (SIXTEEN % 8 == 0)? (SIXTEEN / 8) : ((SIXTEEN / 8) + 1) int main(); int main() { printf("max = %d\n", TWO); int i; for (i = 0; i < TWO; i++) { printf("%d\n", i); } return 0; }

This prints:

max = 2 0 1 2 ...

and continues until terminated, When it should be printing simply:

max = 2 0 1

and exiting.

If I do this instead, it works:

#define TWO 2

I thought that this was an issue with the macro's definition... however, if I do the following with the original #define, it seems to work:

... int count = TWO; for (i = 0; i < count; i++) { ...

Can anyone explain what's going on here?

James McNellis · Accepted Answer · 2010-12-15 21:54:29Z

The problem is that the token TWO is replaced by the tokens with which you defined the macro, so this:

i < TWO

becomes this:

i < (SIXTEEN % 8 == 0)? (SIXTEEN / 8) : ((SIXTEEN / 8) + 1)

Because of operator precedence, this is read as:

(i < (SIXTEEN % 8 == 0)) ? (SIXTEEN / 8) : ((SIXTEEN / 8) + 1)

You need extra parentheses so that when TWO is replaced by its replacement list, you get the result you want:

#define TWO ((SIXTEEN % 8 == 0)? (SIXTEEN / 8) : ((SIXTEEN / 8) + 1)) ^ ^

When using macros, it's best to use parentheses wherever you can to ensure the result is what you expect.

+1 The general rule of thumb is: if your macro is supposed to expand into an expression then make sure it's enclosed in a (matching) pair of parens. The one exception is if it's guaranteed to expand into a single token (like the OP's SIXTEEN macro).

Blagovest Buyukliev · Accepted Answer · 2010-12-15 22:31:38Z

Always enclose macros in parentheses because you don't always know the context in which it will be used - it could happen that an adjacent operator has a higher precedence and the macro will not evaluate correctly.

Not using parentheses is only justified if the #define'd symbol is a single token, such as 5 or "hello world".

If you consider calling your macro with arguments that are expressions and not just single tokens, enclose every occurrence of that argument in parentheses for the same reason as above.

Another thing to avoid is passing expressions that have side effects as macro arguments. If the corresponding macro argument is referenced more than once in its definition, the evaluation will be performed more than once, and that usually isn't what's desired.

Chris Lutz · Accepted Answer · 2010-12-15 21:55:07Z

Expand the macro, and look at your for loop after the macro expansion:

for (i = 0; i < (16 % 8 == 0)? (16 / 8) : ((16 / 8) + 1); i++)

See? i < (16 % 8 == 0) is the condition of the ?: operator. You need to put a pair of parenthesis around the definition of TWO.

Collectives™ on Stack Overflow

Weird macro definition issue

3 Answers 3

1 Comment

Comments

Comments

Linked

Hot Network Questions

Collectives™ on Stack Overflow

3 Answers 3

1 Comment

Comments

Comments

Linked

Related