Using a constexpr static member of a reference as template argument

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, 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?

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
– Evg, Commented Aug 6, 2018 at 15:00
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
– Marcel, Commented Aug 6, 2018 at 15:03

T.C. · Accepted Answer · 2018-08-06 14:57:27Z

10

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;

[...]

answered Aug 6, 2018 at 14:57

T.C.

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5 Comments

Lukas Barth Over a year ago

Thanks, but shouldn't the … unless […] it is initialized with a constant expression … part be true here? Also, should the expression br.b really evaluate to an id-expression? It's a static member access, thus the object itself is irrelevant?

T.C. Over a year ago

b is not a constant expression. The object expression is still evaluated even though the result is discarded.

Jarod42 Over a year ago

Sad that it is not interpreted as BoolHolder::b first.

T.C. Over a year ago

@Jarod42 Would you prefer f().x to not evaluate f() if x is a static data member?

David G Over a year ago

What does "unless the reference has a preceding initialization" mean?

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