Question

I am compiling the following simple program with g++-4.6.1 --std=c++0x:

#include <algorithm>

struct S
{
    static constexpr int X = 10;
};

int main()
{
    return std::min(S::X, 0);
};

I get the following linker error:

/tmp/ccBj7UBt.o: In function `main':
scratch.cpp:(.text+0x17): undefined reference to `S::X'
collect2: ld returned 1 exit status

I realize that inline-defined static members do not have symbols defined, but I was under the (probably flawed) impression that using constexpr told the compiler to always treat the symbol as an expression; so, the compiler would know that it is not legal to pass a reference to the symbol S::X (for the same reason you can’t take a reference to the literal 10).

However if S is declared as namespace, i.e. “namespace S” instead of “struct S”, everything links fine.

Is this a g++ bug or do I still have to use a trick to workaround this annoyance?

Answer 1

I don’t think this is a bug. If you change the constexpr to const, it still fails, with the exact same error.

You’ve declared S::X, but not defined it anywhere, so there’s no storage for it. If you do anything with it that needs to know the address of it then you’ll need to define it somewhere also.

Examples:

int main() {
      int i = S::X; // fine
      foo<S::X>(); // fine
      const int *p = &S::X; // needs definition
      return std::min(S::X, 0); // needs it also
}

The reason for this is that constexpr can be evaluated at compile time, but it’s not required to be evaluated as such, and can equally happen at runtime. It doesn’t instruct “the compiler to always treat the symbol as an expression”, it hints that it would be sensible and permissible to do so if the compiler felt like it.