如何使用完全限定的名称继承基类的构造函数？

Question

Suppose I have the following class:

class foo : public std::runtime_error
{};

To bring in the constructors of std::runtime_error into my class, you'd think I could do this:

using std::runtime_error::std::runtime_error;

However, this doesn't work. Normally I'd think to do this:

using super = std::runtime_error;
using super::super;

This does work, but not sure if this is a workaround or the only real solution. Upon testing with Clang, this works:

using runtime_error::runtime_error;

However I'm curious why this works since it isn't fully qualified. To throw a wrench in the mix, I tried breaking it by introducing multiple inheritance:

namespace n
{
    class runtime_error {};
}

class foo : public std::runtime_error, public n::runtime_error
{
public:
    using runtime_error::runtime_error;
};

Now using runtime_error::runtime_error doesn't work anymore, because it's ambiguous. However, the super alias still works (since you explicitly pick the base class you want to inherit from), but could get nasty if you have to define aliases for each inherited type (eg super1 , super2 ).

What is the proper way to inherit constructors from a base class that is in a different namespace, when fully-qualified name syntax (using :: ) is ambiguous with the :: used to specify the member constructor?

Answer 1

The reason why you have a compilation error when trying to use using std::runtime_exception::std::runtime_exception in the class is because you need to provide name of the constructor. And the name of the constructor for std::runtime_exception is not std::runtime_exception , it is runtime_exception - the name of the constructor is the unqualified type. Because of that, correct notation would be

using std::runtime_exception::runtime_exception;

Your other solution with type alias work as well, and although it might seem confusing at first - why does it work with type alias, when type alias is indeed std::runtime_exception - it is helpful to remember that type aliasing is not textual substitution, and thus it works as expected.

Both approaches are valid and which one to choose is totally dependent on you and your code.