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As the title says, I'm wondering if there's a way to expose a protected constructor in a derived class (that is, to change access from protected to public). Consider the following (contrived) example:

struct A {
    int a;
    int b;
protected:
    A(int, int) {};
    void do_something();
};

struct B : A {
    B() : A(0, 0) {};
    using A::A;
    using A::do_something;
};

int main() {
    B b(0, 0); // error: 'A::A(int, int)' is protected
    //B b{}; // no errors
    b.do_something();
}

So one can change the access of protected member functions, but not of constructors? If so, what is the rationale for the restriction?

Workaround: Variadic template with argument forwarding can serve as a workaround that would perform identically.

Speculation for the rationale: by default the constructor of base classes are not "inherited" in the regular sense; the constructor of derived class must construct the base class instance(s) first. When using is used upon regular member functions, it introduces the functions into the current declarative region thus changing access; on constructors, using only brings base constructor to the "normal inheritance" level (identical to other member functions without using). Because only in some cases is constructor inheritance (reuse of base constructors) useful, the standard committee people designated using X::X syntax for constructor inheritance rather than the stronger namespace teleporting.

YiFei
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    [This](https://stackoverflow.com/questions/21015909/c11-inheriting-constructors-and-access-modifiers) points to why it doesn't work, but there is no answer for why this behavior was chosen. – NathanOliver Jun 12 '20 at 19:52
  • Thanks NathanOliver for the link. For people who use new C++ standards (I'm using n4835), the relevant section is [namespace.udecl]/1, which reads: If the using-declarator names a constructor, it declares that the class inherits the set of constructor declarations introduced by the using-declarator from the nominated base class. [So it doesn't change the declarative region and thereby access control] – YiFei Jun 12 '20 at 20:10
  • @Jarod42, always use `-pedantic` :) – YiFei Jun 12 '20 at 20:10

1 Answers1

1

What you are trying to do is to call the constructor B, with two parameters, but it doesn't take any.

The error you get, is about the constructor of B, being protected, but if you look carefully, that is not the constructor you declared:

error: 'B::B(int, int)' is protected

Why is that? Because the compiler is trying to resolve your call to something valid, and it finds the base class constructor to match the signature. But it is still protected so it fails.

What you need to do, to "relaunch" the protected constructor to an inherited class is to provide a matching constructor:

struct B : A {
    B() : A(0, 0) {};
    B(int a, int b) : A(a, b) {};
    using A::A;
    using A::do_something;
};

int main() {
    B b;
    B b2(1, 2);
    //b.do_something();
}

http://cpp.sh/9tvik

The reason why, you cannot change the accessor level of a constructor the same way you change it for a method, is most probably to be searched into the non-virtual nature of the constructors. [citation needed :)]

Jarod42
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crsn
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