How typedef works for function pointers

Question

I think I may be suffering from the dreaded \"accidental programmer\" disease, at least when it comes to typedefs and function pointers. So I\'ve been experimenting with all

Answer 1

It turns out that, in C/C++, both funcname and &funcname will yield the address of funcname and can be assigned to a function pointer variable. This is actually just an oddity of how the syntax was designed for the language(s).

Answer 2

The thing about function pointers is that they're function pointers! :-) This is how you get your third sample to work:

#include <stdio.h>
typedef void (*print)(void);
//            ^
void do_something (void) { printf("Hello World\n"); }
int main (void) {
    print pr;
    pr = do_something; // &do_something would also work.
    pr();
    return 0;
}

In terms of whether you use funcName or &funcName, it doesn't matter (in C at least). Section 6.3.2.1 Lvalues, arrays and function designators states:

A function designator is an expression that has function type. Except when it is the operand of the sizeof operator or the unary & operator, a function designator with type "function returning type" is converted to an expression that has type "pointer to function returning type".

Answer 3

The address of a function name and the plain function name both mean the same thing, so & has no effect on a function name.

Similarly, when using function pointers, multiple dereferencing isn't a problem:

#include <stdio.h>
typedef void print(void);
static void dosomething(void) { printf("Hello World\n"); }

int main(void)
{
    print *f1 = dosomething;
    print *f2 = &dosomething;
    f2();
    (f1)();
    (*f1)();
    (**f2)();
    (***f1)();
    (****f2)();
    (*****f1)();
}

That compiles cleanly under:

gcc -O3 -g -Wall -Wextra -Werror -Wmissing-prototypes -Wstrict-prototypes \
    -Wold-style-definition -std=c99 xx.c -o xx

I would not claim that multiple stars is good style; it isn't. It is 'odd, and (yes, you may say it) perverse'. One is sufficient (and the one star is mainly for people like me who learned to program in C before the standard said "it is OK to call a function via a pointer without using the (*pointer_to_function)(arg1, arg2) notation; you can just write pointer_to_function(arg1, arg2) if you like"). Yes, it is weird. No, no other type (or class of types) exhibits the same behaviour, thank goodness.

Answer 4

Like C, C++ has pointer to functions: void (*)() for example is a pointer to a function that takes no argument and returns no value. However, C++ has also introduced references to functions void (&)() and there are implicit conversions between the two (though I don't remember the rules exactly).

Therefore:

• funcname is a reference to function
• &funcname is a pointer to function

Note that taking the address (or a reference to) a function that is overloaded requires a static_cast to the exact type (to resolve the overload).