[Bug c++/68588] New: GCC requires constexpr for template non-type parameter, even though constexpr conversion operator exists

["petr at kalinin dot nnov.ru" <gcc-bugzilla at gcc dot gnu dot org>]

https://gcc.gnu.org/bugzilla/show_bug.cgi?id=68588

            Bug ID: 68588
           Summary: GCC requires constexpr for template non-type
                    parameter, even though constexpr conversion operator
                    exists
           Product: gcc
           Version: 5.2.0
            Status: UNCONFIRMED
          Severity: normal
          Priority: P3
         Component: c++
          Assignee: unassigned at gcc dot gnu.org
          Reporter: petr at kalinin dot nnov.ru
  Target Milestone: ---

Consider the following code:

struct A {
    constexpr operator int() { return 42; }
};

template <int>
void foo() {}

void bar(A a) {
    foo<a>();
}

int main() {
    foo<A{}>();

    const int i = 42;
    foo<i>();  // (1)

    A a{};

    static_assert(i == a, "");
    bar(a);
    foo<a>();  // error here
}

Clang 3.7 with c++14 accepts this, while gcc 5.2.0 with c++14 (by "g++
--std=c++14 b.cpp -o b") does not, producing the following message:

b.cpp: In function âint main()â:
b.cpp:22:9: error: the value of âaâ is not usable in a constant expression
     foo<a>();  // error here
         ^
b.cpp:18:7: note: âaâ was not declared âconstexprâ
     A a{};
       ^

This behavior of GCC seems at at least strange, as it allows "a" in
static_assert, as well in bar(), but not directly in foo<a>. Moreover,
A::operator int() is declared constexpr and does not evaluate any class
members, therefore use of "a" should be converted constant expression.

----
This is from my question
http://stackoverflow.com/questions/33957274/type-conversion-at-template-non-type-argument-without-constexpr/33958291.
The last sentence here is reworded from Serge Ballesta's answer there.