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Home/ Questions/Q 3439282
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Editorial Team
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Editorial Team
Asked: May 18, 20262026-05-18T08:17:27+00:00 2026-05-18T08:17:27+00:00

class Base { public: void foo(){} }; class Derived : public Base { private:

  • 0
class Base
{
    public: void foo(){}
};

class Derived : public Base
{
    private:
    using Base::foo;
};

int main()
{
   Derived d;
   d.foo();
}

Is the code legal? The declaration using Base::foo is in the private section of the derived class. So the call d.foo() shouldn’t compile, am I right?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-18T08:17:27+00:00Added an answer on May 18, 2026 at 8:17 am

    The Standard in section 11.2/4 says

    A member m is accessible when named in class N if

    — m as a member of N is public, or

    — m as a member of N is private, and the reference occurs in a member or friend of class N, or

    — m as a member of N is protected, and the reference occurs in a member or friend of class N, or in a
    member or friend of a class P derived from N, where m as a member of P is private or protected, or

    — there exists a base class B of N that is accessible at the point of reference, and m is accessible when named in class B.

    However the Standard also says that

    §11.3/1 “The access of a member of a base class can be changed in the derived class.

    In your code the access of the member foo has been changed in the derived class. So the code shouldn’t compile but this is still an active issue with open status So some compilers compile the code (Comeau and Intel C++) whereas g++ and MSVC++ (correctly) reject it.

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