Question

I am confused by the form of parameters for function pointers. The following two:

int fun(int (*g)())
{
    cout << g() << endl;
}

int fun(int g())
{
    cout << g() << endl;
}

Both these two definitions work well. But as you have noticed, there are some differences in the prototypes of these two functions:

• the first one takes parameter int (*g)(),
• while the second takes parameter int g().

My question is are there any difference between them?

Answer 1

In the second case, the function type adjusts to become pointer-to-function-type, which makes the both function identical.

int fun(int (*g)()); 
int fun(int g()); //same as above, after type adjustment

The C++03 Standard says in §13.1/3,

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())
]

---

It’s same as with array to pointer adjustement, with which we’re more familiar:

int fun(int *a); 
int fun(int a[]);   //same as above, after type adjustment
int fun(int a[10]); //same as above, after type adjustment

All are same!

You can a detail answer by me here:

• Reference to Function syntax – with and without &