pointer to const vs usual pointer (for functions)

Question

Is there any difference between pointer to const and usual pointer for functions? When it is suitable to use const qualifier for stand alone functions?

I wrote short sample to illustrate my question:

#include <iostream>
using namespace std;

int sum( int x, int y ) { return x + y; }
typedef int sum_func( int, int );

int main()
{
    const sum_func* sum_func_cptr = &sum; // const function
    sum_func* sum_func_ptr = &sum;        // non-const function ?

    // What is the difference between sum_func_cptr and sum_func_ptr

    int x = sum_func_cptr( 2, 2 );
    cout << x << endl;

    int y = sum_func_ptr( 2, 2 );
    cout << y << endl;

    sum_func_cptr = 0;
    sum_func_ptr = 0;

    return 0;
}

g++ gives no warnings. That's why I ask.

Answer 1

Your code is ill-formed with regard to C++03. You can not ever construct a const (or volatile) qualified function type. Whenever you do, your program becomes ill-formed.

This rule has been changed for C++1x, to make the compiler ignore the const / volatile. C++ compilers will usually already implement this rule even in C++03 mode. Thus, the following two will define the same function twice, and results in a compilation error.

typedef void Ft();


void f(Ft const*) { }
void f(Ft *) { } // another definition!

Here is the proof of my claim. C++03, 8.3.5/1

A cv-qualifier-seq shall only be part of the function type for a nonstatic member function, the function type to which a pointer to member refers, or the top-level function type of a function typedef declaration. The effect of a cv-qualifier-seq in a function declarator is not the same as adding cv-qualification on top of the function type, i.e., it does not create a cv-qualified function type. In fact, if at any time in the determination of a type a cv-qualified function type is formed, the program is ill-formed.

Here is that text for C++1x, 8.3.5/7 n2914:

A cv-qualifier-seq shall only be part of the function type for a non-static member function, the function type to which a pointer to member refers, or the top-level function type of a function typedef declaration. The effect of a cv-qualifier-seq in a function declarator is not the same as adding cv-qualification on top of the function type. In the latter case, the cv-qualifiers are ignored.

The above says that the below is valid, though, and creates the function type for a function that can declare a const member function.

typedef void Ft() const;
struct X { Ft cMemFn; };
void X::cMemFn() const { }

Answer 2

Stand alone functions are const by definition. Hence there is no difference between a const and a non-const function pointer.

Answer 3

I think you meant,
sum_func* const sum_func_cptr instead of const sum_func* sum_func_cptr.

sum_func* const sum_func_cptr = ∑
sum_func* const sum_func_cptr = &sum_new; // will not compile.
// whereas,
const sum_func* sum_func_cptr = ∑  // will compile
const sum_func* sum_func_cptr = &sum_new; // will compile.
sum_func* sum_func_cptr = ∑  // will compile
sum_func* sum_func_cptr = &sum_new; // will compile.

-Jagannath.

Answer 4

As an interesting aside, the const specifier does not seem to have an effect even when used on pointers to member functions.

#include <iostream>
using namespace std;

class Foo {
public:
  int sum( int x, int y ) {
    _x = x;
    _y = y;
    return x + y;
  }
private:
  int _x;
  int _y;
};

typedef int (Foo::*sum_func)(int,int);

int main()
{
    Foo f;
    const sum_func sum_func_cptr = &Foo::sum; // const pointer
    sum_func sum_func_ptr = &Foo::sum;        // non-const pointer

    int x = (f.*sum_func_cptr)( 2, 2 );
    cout << x << endl;

    int y = (f.*sum_func_ptr)( 2, 2 );
    cout << y << endl;

    const sum_func* sum_func_cptr_cptr = &sum_func_cptr;
    sum_func* sum_func_ptr_ptr = &sum_func_ptr;

    x = (f.**sum_func_cptr_cptr)( 2, 2 );
    cout << x << endl;

    y = (f.**sum_func_ptr_ptr)( 2, 2 );
    cout << y << endl;

    return 0;
}

Answer 5

I think there has been a basic misunderstanding in the previous replies.

    const sum_func sum_func_cptr = &Foo::sum; // const pointer

That means that sum_func_cptr is a constant pointer to a function, that is you can initialize it with a non-const member function, but you can't change it later to point to another function, because const refers to the variable. That's equivalent to:

    sum_func const sum_func_cptr = &Foo::sum; // const pointer

Don't you agree? :-)

-Paolo