Re: ambiguous access of 'Base' ??

David Wilkinson <>
Wed, 09 Apr 2008 17:32:31 -0400
Victor Bazarov wrote:

You need to tell the compiler which of the two 'Base' types you
want created. You need to say

    return Derived1::Base();


    return Derived2::Base();

Alternatively, you can make 'Base' a virtual base class if you
define *both* Derived1 and Derived2 as

    class Derived1 : public virtual Base

    class Derived2 : public virtual Base

in which case 'Derived' will have only one subobject of type Base
in it. Read up on virtual inheritance.

int main(int argc,char *argv[])
Derived d;

return 0;

cl -W4 -nologo /J aa.cpp user32.lib

when i comment out the ":public Base" at "Derived2" then the code
compiles and the result is :
0012FF70 : 12 : 0
0012FF70 : 29 : 1
0012FF74 : 47 : 2
0012FF70 : 66 : 3
0012FF6C : 12 : 4

so, as expected, Derived::Derived1::Base has nothing to do with the
call to Base(), Base() creates a new item.

I don't understand that statement, sorry.


He means that his make2() method returns a new Base object; it is not returning

I never use this diamond pattern, but I have to say I do not understand this
error either.

David Wilkinson
Visual C++ MVP

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