Using a constexpr static member of a reference as template argument

8
votes

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, b.b is a valid constexpr (or the first Foo<b.b> 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?

1
c++language-lawyerc++17
FWIW, MSVS and icc also compile. – NathanOliver
I think this is related to the question: with constexpr int f(int) { return 0;}, would f(x) be a constexpr for int x;? stackoverflow.com/questions/47696686/… – Evg
I would not bet that the reference is not relevant here. bris not constexpr and so everything based on br is also not constexpr. Maybe as workaround you could use BoolHolder::b, if BoolHolder is not constant within this context you might use decltype. – Marcel

1 Answers

10
votes

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;
  • [...]