Use of 'using' keyword to make inherited constructor public
c++, c++11, inheritance, testing, using
Solution
The Standard explicitly states that inherited constructors retain their access levels:
12.9 Inheriting constructors [class.inhctor]
1 A using-declaration (7.3.3) that names a constructor implicitly declares a set of inheriting constructors. The candidate set of inherited constructors from the class `X` named in the using-declaration consists of actual constructors and notional constructors that result from the transformation of defaulted parameters as follows:
[list of cases omitted]
4 A constructor so declared has the same access as the corresponding constructor in X. It is deleted if the corresponding constructor in `X` is deleted (8.4).
You can call it directly of course:
TestableFoo(int i) : Foo(i) { }
Problem
I am trying to test protected methods and constructors of my class. For this purpose, I tried to subclass it, and re-export its members as public with C++11 `using` keyword: ``` class Foo { protected: Foo(int i) {} void run() {} }; class TestableFoo : public Foo { public: using Foo::Foo; using Foo::run; }; int main() { TestableFoo foo(7); foo.run(); } ``` However, both g++ and clang++ fail to compile it, producing the following error: ``` test.cpp:13:15: error: ‘TestableFoo::TestableFoo(int)’ is protected using Foo::Foo; ^ test.cpp:18:16: error: within this context TestableFoo foo(7); ^ ``` TestableFoo constructor is still protected, even though `run` method becomes public (I confirmed it separately). Why is that so? I could understand either decision (inheriting vs. overwriting visibility), but why is there an inconsistency between methods and constructors?