Question

I have a function pointer inside a struct that gets dynamically set at runtime to the address of another function in various places in my code. It is defined in my header file like this:

    void *(*run)();

During compile time, I get the following warning about this:

    warning: function declaration isn't a prototype

This warning is benign, because the pointer is used in many places in my code to call the function it points to, and everything works just fine. However, I would really like to silence the warning.

If I change it to this:

    void *(*run)(void);

I get compile errors whever I use it, because the various functions that make use of the pointer have different numbers of arguments, and saying void inside the parenthesies tells the compiler it accepts no arguments.

I can’t use a va_list or anything fancy like that, as this is simply a pointer to another function, and I use a single pointer for them all because it keeps the code clean and simple.

I can silence the warning with adding this to my compiler flags:

    -Wno-strict-prototypes

But I’d rather not have to disable compiler warnings with flags if I can avoid it.

So my question is: How do I notate this function pointer in the code in such a way that the compiler is satisfied with the fact that it accepts any number of any kind of arguments?

The code works perfectly. I just want the warning to go away.

Answer 1

Store the pointer as a void * and cast to the appropriate function pointer type when necessary? Keep in mind that it isn’t necessarily safe to call one type of function pointer as if it were another type, so the warning you’re starting out with isn’t entirely invalid.

You can cast a function pointer like so:

void *genericPointer = ...;
void (*fp)(int, int) = genericPointer;
fp(123, 456);

Note that:

• There’s no explicit casting necessary here, as void * can always be cast to any pointer type.
• The initial “void” before (*fp) is the return type of the function pointer.