Comments on: Strong types: inheriting the underlying type’s functionalities

By: Peter

Peter — Mon, 17 Dec 2018 10:28:00 +0000

Hey Jonathan,
First of all, great series, I learn a lot from the these posts.
I found though that the ‘public’ inheritance modifier is missing when the NamedType inherits explicity from the Addable class. It is not possible to call the operator+. (when the variadic template is used, the modifier is there)

By: Bárdosi Péter

Bárdosi Péter — Thu, 08 Jun 2017 04:44:00 +0000

In reply to Jonathan Boccara. OK! Thanks!

By: Jonathan Boccara

Jonathan Boccara — Wed, 07 Jun 2017 20:05:00 +0000

In reply to Bárdosi Péter. Hey Péter, This is just a technical artifact for allowing SFINAE to kick in. Indeed, when instantiating the constructor, T is no longer a template parameter because it has always been resolved earlier, when instantiating the class. Hence we need a new template parameter, T_. And since we want to do the check on T, we set T_=T. Makes more sense? This was done in passing strong types by references, and Jonathan Müller has written a very good post about this technique.

By: Bárdosi Péter

Bárdosi Péter — Wed, 31 May 2017 20:06:00 +0000

In reply to Bárdosi Péter. Maybe T&& is an rvalue reference, but T_&& is a forwarding reference?

By: Bárdosi Péter

Bárdosi Péter — Wed, 31 May 2017 19:50:00 +0000

Hi,
sorry, but I don’t understand this:

template

explicit NamedTypeImpl(T&& value, typename std::enable_if::type = nullptr) : value_(std::move(value)) {}

When I change values type from T&& to T_&&, the code functions differently. Why?

P00t