Which other language has interfaces specifying field access, that does not go th...

siknad · on May 15, 2022

> given that there isn’t even any clean way to specify interface expected by C++ template, all you have is type traits

Concepts?

xyzzyz · on May 15, 2022

Ah, sorry, I was not up to speed with C++, it’s called concepts now.

So, how do I specify, using concepts, that my template expects a type which has a field “.foo” of type int?

siknad · on May 15, 2022

  template<typename T>
  concept has_foo = requires(T x){
  { x.foo } -> std::same_as<int>;
  };

  template<has_foo T>
  int get_foo(T x){...}

  // or

  template<typename T>
  requires has_foo<T>
  void f(T x){...}

  // or even

  void f(has_foo auto x) {...}

BobbyJo · on May 15, 2022

Does the above provide more safety/guarantees/something than just:

type hasFoo interface {

  getfoo() int

  setfoo(int)

}

func f(hasFoo) {}

?

throwaway894345 · on May 16, 2022

The equivalent Go would be something like:

    func f[T interface{.Foo}](T) {}

Of course, this doesn't exist in Go, so we must do:

    func f[T interface{GetFoo(); SetFoo()}](T) {}

And then define setters and getters on every type we want to use, including defining new types (with the commensurate setters/getters) for types that we don't own (i.e., if you import a struct from another package and it doesn't have setters and getters defined, you have to create a new type that implements those setters/getters).

BenFrantzDale · on May 15, 2022

I don’t know Go, but I think less in that concepts are syntax guarantees, not semantic guarantees. I suspect interfaces (assuming they are opt-in) are semantic guarantees. That said, I’ve never run into this distinction being a problem, and you could provide opt-in mechanisms in C++. Of course, the C++ interface has zero runtime cost.

Gibbon1 · on May 16, 2022

Unsure what's going on under the hood but C# allows access to fields transparently to the caller. One of the nicer features of the language.

  foo.baz = 1;        // baz could be a int or could be a setter.
  int baz = foo.baz;  // same