C: macro produced warning "statement with no effect"

Question

I have following code in C:

int do_something(void);

#ifdef SOMETHING
#define DO_SOMETHING() do_something()
#else
#define DO_SOMETHING() 0
#endif

This code produced warning "statement with no effect" when compiled without SOMETHING defined. I am trying to fix it, but there is one problem - code which uses this macro sometimes checks that "return value" and sometimes ignores it. Because of this I cannot use the easiest solution - casting to void in macro itself.

Is it possible to write macro which allows to compare "returned value" and does not produce this warning when it is ignored?

I use gcc to compile my code.

The classic way to do nothing is `#define DO_SOMETHING() ((void)0)`. The cast stops the compiler complaining about 'statement with no effect'. However, if you actually sometimes use the result (which you should since `do_something()` returns an `int`), then that doesn't help as much as a dummy function. See also [C `#define` macro for debug printing](http://stackoverflow.com/questions/1644868/). — Jonathan Leffler, Apr 10 '15 at 17:42

score 2 · Accepted Answer · answered Apr 10 '15 at 15:59

2

Three possible solutions:

Define a do_nothing function, which will get optimized out by gcc:

int do_something(void);
int do_nothing(void) { return 0; } 

#ifdef SOMETHING
#define DO_SOMETHING() do_something()
#else
#define DO_SOMETHING() do_nothing()
#endif

Or, modify the do_something implementation to move the #ifdef check there

int do_something(void)
{
  #ifndef SOMETHING
  return 0;
  #endif

// Your implementation here
}

You can also ignore the warning using #pragma directives.

By the way, which version of gcc and with which flags are you compiling? gcc -Wall -pedantic with GCC 4.9 doesn't produce the warning.

answered Apr 10 '15 at 15:59

gjulianm

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7
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Ahh, I forgot that I can use another function which always returns zero. Thanks!. I use gcc 4.4.7, it produces this warning with -Wall or -Wunused-value. – Daniel Frużyński Apr 10 '15 at 16:56
2

I would define `static inline int do_nothing() { return 0; };` – Basile Starynkevitch Apr 10 '15 at 17:33

score 0 · Answer 2 · edited Apr 10 '15 at 17:40

0

#include <stdio.h>

int do_something(void);

#ifdef SOMETHING
#define DO_SOMETHING() do_something()
#define DO_SOMETHING_ASSIGN() do_something()
#else
#define DO_SOMETHING() 
#define DO_SOMETHING_ASSIGN() 0
#endif

int do_something(void)
{
    static int cnt=0;
    /* code for do something here */
    ++cnt;
    printf("do_something has been called %d %s!\n",cnt,
        (cnt==1)?"time":"times");   
    return cnt;
}

void main(void)
{
    int x;

    DO_SOMETHING();

    x=DO_SOMETHING_ASSIGN();    

    printf("%d %d\n",x,DO_SOMETHING_ASSIGN());

    puts("end!!!");
}

I hope this is usefull to you!

If you save this code as main.c, you may compile it with:

gcc main.c -o main

when you run main, you obtain the following output:

0 0
end!!!

If you compile it with:

gcc main.c -DSOMETHING -o main

When you run main, you obtain the following output:

do_something has been called 1 time!
do_something has been called 2 times!
do_something has been called 3 times!
2 3
end!!!

edited Apr 10 '15 at 17:40

Jonathan Leffler

730,956
141
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answered Apr 10 '15 at 16:41

Sir Jo Black

2,024
2
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There is one problem with this - I would have to modify my huge code base. I need something what can be applied in one file. – Daniel Frużyński Apr 10 '15 at 17:00
I image that you want use/don't use a function that returns a value! I think you use in your code the macro DO_SOMETHING(), but this may be called directly (without assigning the result) or assigning it's returned value. If this is the problem the better way might be to modify the function do_something as gjulianm indicates. – Sir Jo Black Apr 10 '15 at 17:30
int do_something(void) { #ifndef SOMETHING return 0; #else // Your implementation here } #endif – Sir Jo Black Apr 10 '15 at 17:31

C: macro produced warning "statement with no effect"

2 Answers2