Suggested resolution:

Change in 6.8.1 [term.layout.compatible.type] paragraph 10 as follows:

A type is a literal type if it is:

• ...
• a possibly cv-qualified class type (Clause 11 [class]) that has all of the following properties:
  • it has a constexpr destructor (9.2.6 [dcl.constexpr]),
  • it is either
    • is a closure type (7.5.5.2 [expr.prim.lambda.closure]),
    • is an aggregate type (9.4.2 [dcl.init.aggr]) for which that type (if it is a union) or each of its anonymous union members (otherwise) either has at least one variant member of non-volatile literal type or has no variant members, or
    • has at least one constexpr constructor or constructor template (possibly inherited (9.9 [namespace.udecl]) from a base class) that is not a copy or move constructor, and
  • if it is a union, at least one of its non-static data members is of non-volatile literal type, and
  • if it is not a union, all of its non-static non-variant data members and base classes are of non-volatile literal types.

According to 6.8.1 [term.layout.compatible.type] paragraph 10, a type is a literal type only if it satisfies the following:

A type is a literal type if it is:

• ...
• a possibly cv-qualified class type (Clause 11 [class]) that has all of the following properties:
  • it has a constexpr destructor (9.2.6 [dcl.constexpr]),
  • it is either a closure type (7.5.5.2 [expr.prim.lambda.closure]), an aggregate type (9.4.2 [dcl.init.aggr]), or has at least one constexpr constructor or constructor template (possibly inherited (9.9 [namespace.udecl]) from a base class) that is not a copy or move constructor,
  • if it is a union, at least one of its non-static data members is of non-volatile literal type, and
  • if it is not a union, all of its non-static data members and base classes are of non-volatile literal types.

[Note 4: A literal type is one for which it might be possible to create an object within a constant expression. ... —end note]

However, the normative rule disagrees with the note. Consider:

  struct A { A(); };
  union U {
    A a;
    constexpr U() {}
    constexpr ~U() {}
  };

It is certainly possible to create an object of type U in a constant expression, even though U is not a literal type.

In the suggested resolution, the aggregate type rule is intended to capture the fact that it is not possible for aggregate initialization of a non-empty union to leave no active member (and similarly for each anonymous union member in a non-union union-like class).