static assert that template typename T is NOT complete?

Question

Is there a way to static_assert that a type T is Not complete at that point in a header? The idea is to have a compile error if someone adds #includes down

Answer 1

Passing a reference through ... doesn't work.

5.2.2/7:

When there is no parameter for a given argument, the argument is passed in such a way that the receiving function can obtain the value of the argument by invoking va_arg (18.10). [note skipped — n.m.] The lvalue-to-rvalue (4.1), array-to-pointer (4.2), and function-to-pointer (4.3) standard conversions are performed on the argument expression. An argument that has (possibly cv-qualified) type std::nullptr_t is converted to type void* (4.10). After these conversions, if the argument does not have arithmetic, enumeration, pointer, pointer to member, or class type, the program is ill-formed.

Here's a kind-of-working solution hastily adapted from @chris's comment:

#include 
#include 
namespace
{

    template
        constexpr auto is_complete(int) -> decltype(sizeof(T),bool{}) {
            return true;
        }

    template
        constexpr auto is_complete(...) -> bool {
            return false;
        }
}


#define IS_COMPLETE(T) is_complete(0) // or use __COUNTER__ if supported

struct S;

static_assert(IS_COMPLETE(int), "oops 1!");
static_assert(!IS_COMPLETE(S), "oops 2!");

struct S {};

static_assert(IS_COMPLETE(S), "oops 3!");