Why can I access a struct defined inside a function from outside the function through return type deduction?

Question

I was watching one of Jason Turner's videos and I saw that you can define a type inside a function scope and have it available outside of that scope through function return type deduction.

auto f()
{
    struct MyStruct
    {
        int n;
    };
    return MyStruct{};
}

int main()
{
   auto a = f().n;
   return a;
}

Why is this allowed? Is there a paragraph in the C++ 14 standard that allows this?

When trying to get the typeid of MyStruct with clang in compile explorer I saw in the assembly output the type displayed as f()::MyStruct, so there is a scope, but somehow I can access MyStruct outside of that scope. Is this some kind of ADL thing?

The name has a scope, not the type. You don't use the name outside its scope. — Baum mit Augen, Feb 21 '17 at 09:11
There's a duplicate... somewhere. But hey, this one looks nice. — Quentin, Feb 21 '17 at 09:15
Something like [this](http://stackoverflow.com/questions/13532784/why-can-i-use-auto-on-a-private-type)? — smkanadl, Feb 22 '17 at 07:32

Angew is no longer proud of SO · Accepted Answer · 2017-02-21T10:01:07.943

No, there's no ADL involved. Since your translation unit contains the definition of the structure, there's no problem in accessing its members.

The important point is that types don't really exist in scopes: names do. And notice that there's no way for you to use the identifier MyStruct instead of auto when declaring a. The name is as inaccessible as it should be. However, as long as you can get at the type without using the inaccessible name, all is fine.

In principle, this is hardly different from using a private member type:

class X
{
  struct Hidden
  {
    int i;
  };

public:
  Hidden get() const { return {42}; }
};

int main()
{
  X x;
  auto h = x.get();
  assert(h.i == 42);
}

Thank you for clarifying the difference between the type name and the actual type for me ^^. — bluespeck, Feb 21 '17 at 09:45

Why can I access a struct defined inside a function from outside the function through return type deduction?

1 Answers1