Multiple inheritance from multiple inheritance

Question

#include<iostream>
using namespace std;

class base1{
public:
    int x;
};
class base2{
public:
    int x;
};

class derived1: public base1, public base2{
    // Contains...  base1::x, base2::x
};
class derived2: public base1, public base2{
    // Contains...  base1::x, base2::x 
};

class derived: public derived1, public derived2{

};

If I am right, the class derived will contain four integers. But I can't access them with
derived1::base1::x, derived1::base2::x, derived2::base1::x, derived2::base2::x

It shows ambiguity error. ('base1' is ambiguous base of 'derived')
Am I missing something? How should I resolve this?

en.wikipedia.org/wiki/Multiple_inheritance#The_diamond_problem — meowgoesthedog
– meowgoesthedog, Commented Mar 30, 2018 at 21:01

StoryTeller - Unslander Monica · Accepted Answer · 2018-03-30 21:05:16Z

2

Well, when you use the scope resolution operator to disambiguate between members, you don't name a "path" to the member. You instead name, however indirectly, the base class where the member comes from.

Since the class name is injected into the scope of the class, and as a result, into the scope of any class that derives from it, you are just naming base1 and base2 in a verbose manner. I.e. all the following are equivalent for the member access:

derived d;
d.base1::x;
d.derived1::base1::x;
d.derived1::derived1::base1::base1::x;

That's why the ambiguity occurs. The way to do it, is to cast the object into a derived1& or derived2&, and then use member access via base1::x and base2::x on that.

static_cast<derived1&>(d).base1::x;

answered Mar 30, 2018 at 21:05

StoryTeller - Unslander Monica

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4 Comments

Eljay Over a year ago

Good technique! The one (and only) time I got to speak with Bjarne, I had asked him how one should do that. Now I know.

Symon Saroar Over a year ago

Didn't know that. Thank you very much.

Symon Saroar Over a year ago

what if x is declared protected: in base1 and base2?

StoryTeller - Unslander Monica Over a year ago

@SymonSaroar - Access specifiers don't affect how lookup works. They only make the access allowed or disallowed after the lookup is done.

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