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.

I think the answer is that because ostream::operator<<(ostream&, const complex<T> &) passes by reference, this is considered odr-use of the argument, therefore you need a definition (for the reference to refer to). — M.M
– M.M, Commented Dec 2, 2015 at 21:55
Yes. Without taking any reference, first code works. float or double are passed by value, so the second code works. Thank you. — Titas Chanda
– Titas Chanda, Commented Dec 2, 2015 at 22:07
@vsoftco Thank you for the note. That thread contains some useful explanations. I didn't find it earlier. — Titas Chanda
– Titas Chanda, Commented Dec 3, 2015 at 17:51

David Foerster · Accepted Answer · 2017-10-04 04:20:17Z

19

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.

edited Oct 4, 2017 at 4:20

David Foerster

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answered Dec 2, 2015 at 21:56

Dimitrios Bouzas

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2 Comments

Alba Mendez Over a year ago

for anyone else confused at this: from C++17, due to changes that affect inline variables in general, the namespace definition becomes unnecessary (but still allowed / ignored for backwards compat)

Notinlist Over a year ago

Workaround: create a copy of it just before using template<typename T> T copy(T v){ return v; } It will create a temporary, and a reference to constant can be taken from a temporary. Will be optimized easily.

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