g++ How to get warning on ignoring function return value

邮差的信 提交于 2019-11-27 07:36:16
Arun

Thanks to WhirlWind and paxdiablo for the answer and comment. Here is my attempt to put the pieces together into a complete (?) answer.

-Wunused-result is the relevant gcc option. And it is turned on by default. Quoting from gcc warning options page:

-Wno-unused-result

Do not warn if a caller of a function marked with attribute warn_unused_result (see Variable Attributes) does not use its return value. The default is -Wunused-result

So, the solution is to apply the warn_unused_result attribute on the function.

Here is a full example. The contents of the file unused_result.c

int foo() { return 3; }

int bar() __attribute__((warn_unused_result));
int bar() { return 5; }

int main()
{
    foo();
    bar();    /* line 9 */
    return 0;
}

and corresponding compilation result:

$gcc unused_result.c 
unused_result.c: In function ‘main’:
unused_result.c:9: warning: ignoring return value of ‘bar’, declared with attribute warn_unused_result

Note again that it is not necessary to have -Wunused-result since it is default. One may be tempted to explicitly mention it to communicate the intent. Though that is a noble intent, but after analyzing the situation, my choice, however, would be against that. Because, having -Wunused-result in the compile options may generate a false sense of security/satisfaction which is not true unless the all the functions in the code base are qualified with warn_unused_result.

Since C++17 you can use the [[nodiscard]] attribute.

Example:

[[nodiscard]] int bar() {
  return 42;
}

-Wunused-result should do this for you. This isn't one of the warnings -Wall turns on:

http://gcc.gnu.org/onlinedocs/gcc/Warning-Options.html

The function has to have the warn_unused_result attribute applied to it (Thanks paxdiablo).

The answers about using __attribute__((warn_unused_result)) are correct. GCC isn't so good at this functionality, though! Be aware: it will not warn for non-POD types. That means, for example, if you return a class with a destructor (or a class with instance variables with destructors) you'll never see a warning about ignoring the result.

Relevant bug: https://gcc.gnu.org/bugzilla/show_bug.cgi?id=66177

Example where it fails:

struct Error {
~Error();
};

__attribute__((warn_unused_result)) Error test();

int main()
{
    test();
    return 0;
}

So, don't rely on this for return types which aren't pretty simple.

I solved the problem like this:

#define ignore_result(x) if (x) {}

then instead of (void)foo() use ignore_result(foo())

Then the code compiles with -Wall just fine.

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