It is not clear whether a defaulted consteval function is still an immediate function even if it is not a valid constexpr function. For example:

  template <typename Ty>
  struct A {
    Ty n;
    consteval A() {}
  };

  template <typename Ty>
  struct B {
    Ty n;
    consteval B() = default;
  };

  A<int> a;
  B<int> b;

The declarations of a and b should both fail due to an uninitialized member n in each of A and B. The = default; should not make a difference. However, there is implementation divergence. We should be able to lean on 7.7 [expr.const] bullet 5.5 to handle this when the immediate invocation is required.

Possible resolution:

Change in 9.2.6 [dcl.constexpr] paragraph 7 as follows:

If the instantiated template specialization of a constexpr templated function template or member function of a class template would fail to satisfy the requirements for a constexpr function, that specialization is still a constexpr function, even though a call to such a function cannot appear in a constant expression. Similarly, if the instantiated template specialization of a consteval templated function would fail to satisfy the requirements for a consteval function, that specialization is still an immediate function, even though an immediate invocation would be ill-formed. If no specialization of the template would satisfy the requirements for a constexpr or consteval function when considered as a non-template function, the template is ill-formed, no diagnostic required.