Understanding static constexpr member variables

Question

I have some confusions regarding static constexpr member variables in C++11.

In first.hpp

template<typename T> struct cond_I { static constexpr T value = 0; };    // specialization  template<typename T> struct cond_I< std::complex<T> > { static constexpr std::complex<T> value = {0,1}; };  

In main() function

cout << cond_I<double>::value << endl;            // this works fine cout << cond_I< complex<double> >::value << endl; // linker error 

However if I add the following line to first.hpp everything works fine.

template<typename T1>  constexpr std::complex<T1> cond_I< std::complex<T1> >::value; 

What I understand (I may be wrong) is, that cond_I< std::complex<double> >::value needs a definition, but in the previous case it only has the declaration. But then what about cond_I<double>::value? Why it does not require a definition?

Again, in another header file, second.hpp, I have:

In second.hpp

// empty struct template<typename T> struct eps { };   // special cases template<> struct eps<double> {   static constexpr double value = 1.0e-12; };  template<> struct eps<float> {   static constexpr float value = 1.0e-6; }; 

In this case, following codes works perfectly without any definition of eps<>::value.

In main() function

cout << eps<double>::value << endl;    //  works fine cout << eps<float>::value << endl;     //  works fine 

Can someone please explain me the different behaviors of static constexpr member variables, in these scenarios?

These behaviors are also the same for gcc-5.2 and clang-3.6.

Answer 1

According to the standard 9.4.2/p3 Static data members [class.static.data] (Emphasis Mine):

If a non-volatile const static data member is of integral or enumeration type, its declaration in the class definition can specify a brace-or-equal-initializer in which every initializer-clause that is an assignment-expression is a constant expression (5.20). A static data member of literal type can be declared in the class definition with the constexpr specifier; if so, its declaration shall specify a brace-or-equal-initializer in which every initializer-clause that is an assignment-expression is a constant expression. [ Note: In both these cases, the member may appear in constant expressions. — end note ] The member shall still be defined in a namespace scope if it is odr-used (3.2) in the program and the namespace scope definition shall not contain an initializer.

As M.M earlier explained in the comments ostream::operator<<(ostream&, const complex<T>&) passes by reference so value is considered odr-used in the program. Thus, as the wording above dictates you have to provide a definition.

Now as you’ve already found out fundamental types are passed by value, that it is why no definition required.