CODEWITHSUNDEEP
c++c++11
R. Bob
R. Bob

Reputation: 3

Calling non-const member function pointer from const member function

I have the following code:

class MyClass;
typedef void(MyClass::*func_ptr)();
class MyClass
{
public:

MyClass()
{
    f = &MyFunc1;
}

void MyFunc1()
{
    // ...
}

void MyFunc2() const
{
    (this->*f)();
}

func_ptr f;

};

If I try to compile, it fails because of MyFunc2() which is a const method trying to call the function pointer of type func_ptr which is non-const.

I'm trying to figure out the best way to cast this. I can use a standard C style cast:

    typedef void(MyClass::*func_ptr2)() const;
    func_ptr2 f2 = (func_ptr2)f;
    (this->*f2)();

But I'd prefer to use a C++ cast (i.e. static_cast, reinterpret_cast, or const_cast). I got this to compile with reinterpret_cast but I can't get it working with const_cast. I thought const_cast was mean't to be used for this case? Also, is there a cleaner way to do this without having to create another typedef?

Upvotes: 0

Views: 1043

Answers (3)

Florent DUGUET
Florent DUGUET

Reputation: 2916

The following seems to compile:

MyClass()
{
    f = &MyClass::MyFunc1;
}

void MyFunc1()
{
    // ...
}

void MyFunc2() const
{
    MyClass* mc = const_cast<MyClass*>(this);
    (mc->*f)();
}

Upvotes: 0

rems4e
rems4e

Reputation: 3172

The typical solution would be (const_cast<MyClass*>(this)->*f)();.

This is legal as long as the MyClass instance has been created non-const. Else, this would invoke undefined behavior.

Upvotes: 2

Maxim Egorushkin
Maxim Egorushkin

Reputation: 136266

There are a couple of issues in this code. 

 f = &MyFunc1;

Is not well-formed C++ because taking an address of a non-static member function requires a fully qualified name:

 f = &MyClass::MyFunc1;

this is MyClass const* in a const member function. You may need to cast away that const:

(const_cast<MyClass*>(this)->*f)();

A better solution may be to rethink the design to avoid fighting the type system with const_cast.

Upvotes: 1

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