Static constexpr members seem not to go along with std::min [duplicate]

This question already has an answer here:

• Why does constexpr static member (of type class) require a definition? 1 answer

Here is a problem the reason of which is quite obscure to me, but the workaround of which is fortunately quite easy.

Consider the following code (let me call it my main.cpp):

#include <algorithm>

struct Foo {
    static constexpr float BAR = .42;

    float operator()() const noexcept {
        float zero = .0;
        return std::min(zero, BAR);
    }
};

int main() {
    Foo foo;
    foo();
}

When I tried to compile it, I got the error:

foobar:~/stackoverflow$ g++ -std=c++11 main.cpp
/tmp/ccjULTPy.o: In function 'Foo::operator()() const':
main.cpp:(.text._ZNK3FooclEv[_ZNK3FooclEv]+0x1a): undefined reference to `Foo::BAR'
collect2: error: ld returned 1 exit status

The same happens (quite obviously) also if I use the following statement:

return std::min(zero, Foo::BAR);

Below a slightly modified version of the example above.
This one compiles with no error, even though I'm still referring to the BAR member:

#include <algorithm>

struct Foo {
    static constexpr float BAR = .42;

    float operator()() const noexcept {
        float zero = .0;
        float bar = BAR;
        return std::min(zero, bar);
    }
};

int main() {
    Foo foo;
    foo();
}

I didn't succeed in understanding why the latter version compiles fine while the former ends with an error.
As far as I know, both the versions are correct and should compile, but I strongly suspect that I'm missing something important here.

Any suggestion?

Here my compiler's version: g++ (Debian 5.3.1-5) 5.3.1 20160101.