死亡钻石和范围解析运算符（C ++）

Question

I have this code (diamond problem):

#include <iostream>
using namespace std;

struct Top
{
    void print() { cout << "Top::print()" << endl; }
};

struct Right : Top 
{
    void print() { cout << "Right::print()" << endl; }
};

struct Left : Top 
{
    void print() { cout << "Left::print()" << endl; }
};

struct Bottom: Right, Left{};

int main()
{
    Bottom b;
    b.Right::Top::print();
}

I want to call print() in Top class.

When I try to compile it I get error: 'Top' is an ambiguous base of 'Bottom' on this line: b.Right::Top::print(); Why is it ambiguous? I explicitly specified that I want Top from Right and not from Left .

I don't want to know HOW to do it, yes it can be done with references, virtual inheritance, etc. I just want to know why is b.Right::Top::print(); ambiguous.

Answer 1

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();

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

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