How can an incomplete type be used as a template parameter to vector here?

Question

TIL the following program is legal and whatnot:

#include <vector>

struct Bar;

struct Foo
{
    using BarVec = std::vector<Bar>::size_type;
};

struct Bar {};

int main()
{
   Foo f;
}

How? Bar is an incomplete type so the compiler has no way of knowing what std::vector<Bar> is, or that it contains a member size_type, or that the member size_type is a type.

The only explanation I can come up with is that any hypothetical specialisation would (presumably) have to already be in scope to cause size_type to take on a meaning different from that given in the "base" template definition, and size_type is not a dependent name (both factors contributing to the compiler's certainty).

What's the legal rationale here?

Answer 1

I think in practice this may work but from what I can tell this looks like undefined behavior. From the draft C++11 standard 17.6.4.8 [res.on.functions]:

In particular, the effects are undefined in the following cases:

[...]

• if an incomplete type (3.9) is used as a template argument when instantiating a template component, unless specifically allowed for that component.

Although instantiating a template component does not seem like a well-defined term.

I came to this via LWG defect 611 which added:

unless specifically allowed for the component.

to the end of the bullet above so it now reads:

if an incomplete type (3.9) is used as a template argument when instantiating a template component, unless specifically allowed for the component.

as an exception for shared_ptr since the above quote conflicted with this quote from 20.6.6.2 [util.smartptr.shared]:

The template parameter T of shared_ptr may be an incomplete type.

Also see N4371: Minimal incomplete type support for standard containers, revision 2.