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

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

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();

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.