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John Trend
John Trend

Reputation: 21

Macro with function parameter which calls the function

I need a macro which simply calls the function which is passed as parameter to that macro. The function is void type with no parameter.

For example, if I have the following function:

void function(void) {
//doSomething
}

I need a macro, APPLY_FUNCTION which is called with the function parameter and simply calls that function:

APPLY_FUNCTION(function);

I've tried to define the macro like:

#define APPLY_FUNCTION(function)    (function)

But it does not work, when I try to call the macro I am getting a "statement with no effect" error.

The question is how should I define that macro?

Upvotes: 0

Views: 842

Answers (2)

Lundin
Lundin

Reputation: 213306

The quick & dirty version is simply this:

#define APPLY_FUNCTION(function) (function)()

where the () on the end means that this is a function call.

I would however prefer to use a type safe C11 version instead:

#define APPLY_FUNCTION(function) _Generic((function), void (*)(void): (function)())

This prevents the user from passing the wrong kind of function to the macro.

Upvotes: 1

Jabberwocky
Jabberwocky

Reputation: 50778

You probably want this:

#include <stdio.h>

#define APPLY_FUNCTION(functionname) functionname()

void function(void) {
  printf("Function called\n");
}

int main()
{
  APPLY_FUNCTION(function);
}

Output:

Function called

With your definition:

#define APPLY_FUNCTION(function)   (function)

The preprocessor generates following line:

(function);

This is syntaxically correct, but it doesn't call the function, hence the warning.

BTW: you should call the macro rather CALL_FUNCTION. You don't apply a function, you call it.

Upvotes: 1

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