Question

#include<stdio.h>
#include<stdlib.h>

int fun1()
{
    printf("I am fun1.");
    return 0;
}

int fun2(int fun())
{
    fun();
    return 0;
}

int main()
{
    fun2(fun1);
    return 0;
}

The above program can run. As far as I am concerned, I can understand int fun2(int (*fun)()), but I do not know how int fun2(int fun()) works. Thank you.

Answer 1

When you write int fun2(int fun()), the parameter int fun() converts into int (*fun)(), it becomes exactly equivalent to this:

int fun2(int (*fun)());

A more famiiar conversion happens in case of array when you declare it as function parameter. For example, if you’ve this:

int f(int a[100]);

Even here the parameter type converts into int*, and it becomes this:

int f(int *a);

The reason why function type and array type converts into function pointer type, and pointer type, respectively, is because the Standard doesn’t allow function and array to be passed to a function, neither can you return function and array from a function. In both cases, they decay into their pointer version.

The C++03 Standard says in §13.1/3 (and it is same in C++11 also),

Parameter declarations that differ only in that one is a function type and the other is a pointer to the same function type are equivalent. That is, the function type is adjusted to become a pointer to function type (8.3.5).

And a more interesting discussion is here:

• Reference to Function syntax – with and without &