Diamond of death and Scope resolution operator (c++)

Question

I have this code (diamond problem):

#include 
using namespace std;

struct Top
{
    void print() { cout << \"Top::print()\" << e         


        

Answer 1

Actually, giving code is working fine as I tried it on Visual Studio 2019. There are two way to solve Diamond Problem; - Using Scope resolution operator - Inherit base class as virtual

Calling print function by b.Right::Top::print() should be executed with no errors. But there is still two objects of your base class (Top) referred from your Bottom class.

You can find additional detail in here

Answer 2

The scope resolution operator is left-associative (though it doesn't allow parentheses).

So whereas you want to refer to A::tell inside B, the id-expression refers to tell inside B::A, which is simply A, which is ambiguous.

The workaround is to first cast to the unambiguous base B, then cast again to A.

Language-lawyering:

[basic.lookup.qual]/1 says,

The name of a class or namespace member or enumerator can be referred to after the :: scope resolution operator applied to a nested-name-specifier that denotes its class, namespace, or enumeration.

The relevant grammar for nested-name-specifier is,

nested-name-specifier:

    type-name ::

    nested-name-specifier identifier ::

So, the first nested-name-specifier is B:: and A is looked up within it. Then B::A is a nested-name-specifier denoting A and tell is looked up within it.

Apparently MSVC accepts the example. Probably it has a nonstandard extension, to resolve ambiguity by backtracking through such specifiers.

Answer 3

Why is it ambiguous? I explicitly specified that I want Top from Right and not from Left.

That was your intent, but that's not what actually happens. Right::Top::print() explicitly names the member function that you want to call, which is &Top::print. But it does not specify on which subobject of b we are calling that member function on. Your code is equivalent conceptually to:

auto print = &Bottom::Right::Top::print;  // ok
(b.*print)();                             // error

The part that selects print is unambiguous. It's the implicit conversion from b to Top that's ambiguous. You'd have to explicitly disambiguate which direction you're going in, by doing something like:

static_cast<Right&>(b).Top::print();