Does the using declaration allow for incomplete types in all cases?

Question

I'm a bit confused about the implications of the using declaration. The keyword implies that a new type is merely declared. This would allow for incomplete types. However, in some cases it is also a definition, no? Compare the following code:

#include <variant>
#include <iostream>

struct box;

using val = std::variant<std::monostate, box, int, char>;

struct box
{
    int a;
    long b;
    double c;
    
    box(std::initializer_list<val>) {
        
    }
};

int main()
{
    std::cout << sizeof(val) << std::endl;
}

In this case I'm defining val to be some instantiation of variant. Is this undefined behaviour? If the using-declaration is in fact a declaration and not a definition, incomplete types such as box would be allowed to instantiate the variant type. However, if it is also a definition, it would be UB no?

For the record, both gcc and clang both create "32" as output.

Answer 1

Since you've not included language-lawyer , I'm attempting a non-lawyer answer.

Why should that be UB?

With a using delcaration, you're just providing a synonym for std::variant<whatever> . That doesn't require an instantiation of the object, nor of the class std::variant , pretty much like a function declaration with a parameter of that class doesn't require it:

void f(val); // just fine

The problem would occur as soon as you give to that function a definition (if val is still incomplete because box is still incomplete):

void f(val) {}

But it's enough just to change val to val& for allowing a definition,

void f(val&) {}

because the compiler doesn't need to know anything else of val than its name.

---

Furthermore, and here I'm really inventing, "incomplete type" means that some definition is lacking at the point it's needed, so I expect you should discover such an issue at compile/link time, and not by being hit by UB. As in, how can the compiler and linker even finish their job succesfully if a definition to do something wasn't found?