使用引用的 constexpr 静态成员作为模板参数

Question

I'm trying to figure out whether GCC or Clang interpret the C++17 standard differently / wrong here.

This is my code, which does compile using GCC 8, but not using Clang 6:

struct BoolHolder {
    constexpr static bool b = true;
};

template<bool b>
class Foo {};

int main() {
    BoolHolder b;
    Foo<b.b> f; // Works

    BoolHolder & br = b;
    Foo<br.b> f2; // Doesn't work
}

I wonder why that is. Obviously, bb is a valid constexpr (or the first Foo<bb> wouldn't be valid). Is br.b not a valid constexpr? Why? The object or the reference itself should have nothing to do with it, since we're accessing a static constexpr member here, right?

If this is really not valid C++17, should the fact that GCC doesn't even warn me (even though I enabled -Wall -Wextra -pedantic ) be considered a bug?

Answer 1

Clang is correct. References are evaluated "eagerly" in constant expressions, so to speak. [expr.const]/2.11:

An expression e is a core constant expression unless the evaluation of e, following the rules of the abstract machine, would evaluate one of the following expressions:

• [...]
• an id-expression that refers to a variable or data member of reference type unless the reference has a preceding initialization and either
  • it is initialized with a constant expression or
  • its lifetime began within the evaluation of e ;
• [...]