When using multiple inheritance, why is this qualified name ambiguous?

Question

I'm trying to access the member variable x in struct Top using a Bottom object.

The code is the following:

#include <cstdio>

struct Top
{
public:
    int x = 1;
};

struct Left : public Top
{
    int x = 2;
};

struct Right : public Top
{
    int x = 3;
};

struct Bottom : public Left, public Right
{
    int x = 4;
}; 

int main()
{
    Bottom b;
    std::printf("value: %d\n", b.Left::Top::x);
    return 0;
}

This gives the following error using gcc 4.8:

main.cpp: In function 'int main()':
main.cpp:27:45: error: 'Top' is an ambiguous base of 'Bottom'
std::printf("value: %d\n", b.Left::Top::x);
                                        ^

How is this ambiguous and how do I access it with a qualified name?

Answer 1

The problem is that C++ has no way to directly express the concept of "multiple-level" class members, such as "the member x of the Top subobject of Left". What Left::Top::x means is "the member x in the type denoted by Left::Top" - and the type denoted by Left::Top is exactly Top.

This is why you can write odd things like

int Left::* ptr = &Right::Top::x;

because the right hand side of the = is exactly equivalent to &Top::x, and a pointer-to-base-class-member is implicitly convertible to a pointer-to-derived-class-member. (The result of this conversion still refers to the member in the base-class subobject of the derived class.)

To disambiguate, you can either do something along the lines of static_cast<Left &>(b).Top::x or use a pointer-to-member - given int Left::* ptr = &Top::x;, b.*ptr will refer to the x in the Top subobject of the Left subobject of b.