How to fix these MIN, MAX macros?

Question

I am getting the following warning:

warning: left-hand operand of comma expression has no effect

The macros are defined below. I am compiling with GCC (4.4.3) on Linux. It is C code.

#define MY_MAX(a,b) \
   ({ __typeof__ (a) _a = (a); \
       __typeof__ (b) _b = (b); \
     _a > _b ? _a : _b; })

#define MY_MIN(a,b) \
   ({ __typeof__ (a) _a = (a); \
       __typeof__ (b) _b = (b); \
     _a < _b ? _a : _b; })

How do I fix them to get rid of the warnings?

[[Update]]

Actually, I found the cause of the warning. It had nothing to do with the macro itself. It was because I was trying to find the min of two numbers, one of which was a #def named as MAXIMUM_ARRAYSIZE. It was defined as:

#define MAXIMUM_ARRAYSIZE (sizeof(size_t)==2,16384,1073741824)

That is a pretty strange macro definition.

Could you show the call site where the warning is triggered? (Tagged GCC, the macros use a GCC extension.) — Mat, Jan 07 '12 at 15:09
Does [this](http://stackoverflow.com/questions/3437404/min-and-max-in-c) help? — César, Jan 07 '12 at 15:12
I don't see a single comma in the code here. So I guess the problem is in the code that uses the macro (which probably contains a comma), not the macro itself. — ugoren, Jan 07 '12 at 15:23

score 1 · Answer 1 · answered Jan 07 '12 at 15:09

1

The error is not in your macro definition.

You'll need to show us the calling code to the MIN/MAX macros. It sounds like you're either trying to use the result of MIN/MAX as incorrectly, or you've passed a token (a or b) containing an unexpected comma.

answered Jan 07 '12 at 15:09

Matt Joiner

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score 0 · Answer 2 · answered Jan 07 '12 at 15:19

I tried the following and it worked fine without any warnings/error with gcc -Wall. Check how you're using those macros!

#include <stdio.h>

#define MY_MAX(a,b) \
   ({ __typeof__ (a) _a = (a); \
       __typeof__ (b) _b = (b); \
     _a > _b ? _a : _b; })

#define MY_MIN(a,b) \
   ({ __typeof__ (a) _a = (a); \
       __typeof__ (b) _b = (b); \
     _a < _b ? _a : _b; })

int main(void)
{
    printf("%d \n", MY_MAX(10,20));
    printf("%d \n", MY_MIN(10,20));
    return 0;
}

Output:

$ gcc ma.c -Wall
$ ./a.out 
20 
10 
$

score -2 · Answer 3 · answered Jan 07 '12 at 15:12

-2

_a < _b ? _a : _b is an expression, not a statement, but you have written it as a statement (all by itself followed by a semicolon).

answered Jan 07 '12 at 15:12

ajd

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3

`({...})` is a GCC extension that *is* an expression (which evaluated to the result of the last "statement"). If it wasn't, you'd get a syntax error instead. And if that was somehow averted, you'd probably get a different warning (e.g. "statement has no effect") - there's no comma operator in the macro. – Jan 07 '12 at 15:16

How to fix these MIN, MAX macros?

3 Answers3