a strange use of typedef

Question

I have never see a grammar in c++ like this before:

typedef int (callback)(int);

what really does this really mean?I just find that if I create a statement

  callback a;

It's effect is very very similar to a forward function declaration.

below is the code I had written

#include<cstdio>

int callbackfunc(int i)
{
    printf("%d\n",i);
    return i*i;
}

// you can also use typedef int (callback)(int) here!
typedef int (*callback)(int);

void func(callback hook)
{
    hook(hook(3));
}

int main()
{
    func(callbackfunc);
    getchar();
        return 0;
}

You can use

typedef int (*callback)(int);//this is very common to use

in this code,but if we change it to

typedef int (callback)(int); //I'm puzzled by this !

this will also get the same result!

and I know typedef int (*callback)(int) and typedef int (callback)(int)
are two completely different stuff.

Answer 1

Its because of the fact that in the parameter declaration, the function-type is adjusted to become a pointer-to-function-type.

typedef int type(int); 
typedef int (*type)(int); 

The first typedef defines a type which is called function-type, while the second typedef defines a type which is called pointer-to-function-type. In the parameter declaration, function-type is adjusted to become a pointer to function type.

§13.1/3 (C++03) says,

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).

[Example:
    void h(int());
    void h(int (*)()); // redeclaration of h(int())
    void h(int x()) { } // definition of h(int())
    void h(int (*x)()) { } // ill-formed: redefinition of h(int())
]

---

An interesting example of the exclusive usage of function-type

Suppose you've a typedef, defined as:

typedef void funtype();

then you can use this to define member-function as:

struct A
{
   //member function declaration. 
    funtype f; //equivalent to : void f();
};

void A::f() //definition
{
  std::cout << "haha" << std::endl;
}

Test code:

int main() {
        A a;
        a.f(); //call member function
}

Output:

haha

Online demo: http://ideone.com/hhkeK

Answer 2

It's because a function implicitly becomes a function pointer where necessary. These are identical:

func(callbackfunc);

func(&callbackfunc);