Making sure the method declaration is inherited

Question

How can I protect from accidental definition of non-inherited method where inherited definition is intended. I am told there is trick to express it, but nobody can recall it

Answer 1

For this exact purpose C++0x introduces the override member function decorator, as is already implemented in VC++ 2005 and later: http://msdn.microsoft.com/en-us/library/41w3sh1c.aspx

Alternatively, VC++ permits the following (presumably compiler-specific):

#include <iostream>

class Base {
public:
    virtual void VeryLongFunctionName(int VeryLongArgumentList) = 0;
};

class C : public Base {
public:
    void Base::VeryLongFunctionName(int VeryLongArgumentList) {
        std::cout << "C::\n";
    }
};

class D : public C {
public:
    void Base::VeryLongFunctionNane(int VeryLongArgumentList) {
    //   ^^^^^^ now causes a compilation error
        std::cout << "D::\n";
    }
};

Answer 2

You have compilation errors -

• int VeryLongFunctionName(int VeryLongArgumentList) supposed to return an int which none of the method definitions is doing so.

• int VeryLongFunctionName(int VeryLongArgumentList) supposed to receive an int.

  p->VeryLongFunctionName(); // Error

With these corrected, you should get the expected results. Check results : http://ideone.com/wIpr9

Answer 3

not exactly what you asked for, but i've used this form to reduce the chance for human error:

class t_very_long_argument_list {
public:
    t_very_long_argument_list(T1& argument1, const T2& argument2);
    /* ... */
    T1& argument1;
    const T2& argument2;
};

int C::VeryLongFunctionName(t_very_long_argument_list& arguments) {
    std::cout << "C::\n";
}