使用'using'关键字将继承的构造函数公开 [英] Use of 'using' keyword to make inherited constructor public

问题描述

我试图测试我的类的受保护的方法和构造函数。为此，我试图将其子类化，然后使用关键字C / C ++ 11 将其成员重新导出为public：

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();
}

$ b <

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构造函数仍然受保护，即使 / code>方法变为public（我单独确认）。为什么会这样呢？我可以理解决策（继承与覆盖可见性），但为什么在方法和构造函数之间存在不一致？

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?

推荐答案

标准明确声明继承的构造函数保留其访问级别：

The Standard explicitly states that inherited constructors retain their access levels:

12.9继承构造函数[class.inhctor]

1隐式命名构造函数的using-declaration（7.3.3）
声明一组继承构造函数。
的候选集合从使用声明
中命名的 X 类中继承的构造函数由实际构造函数和名词构造函数组成，结果为
从默认参数的转换如下：

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:

[已省略的案例列表]

[list of cases omitted]

4 这样声明的构造函数具有与X 中对应的
构造函数相同的访问权限。如果 X
中的相应构造函数已删除（8.4），则删除它。

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).

你可以直接调用它：

TestableFoo(int i) : Foo(i) { }

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