Why static const char * template template member is not initialized

Question

I have some C ++ 11 code which I am trying to connect to Visual C ++ Compiler 2015. The source code works fine, however I need to rewrite it to solve problems using constexpr.

Source code (simplified example)

#include <iostream>

struct String
{
    static constexpr const char * value{ "STRING" };
};

template<typename Base>
class Derived
{
public:
    static constexpr const char * value{ Base::value };
};

template<typename BarType>
struct Foo
{
    static constexpr const char * value{ BarType::value };
};

using Bar = Derived<String>;
using FooBar = Foo<Bar>;

int main()
{
    std::cout << "FooBar::value = " << FooBar::value << std::endl;
}

Fingerprints:

FooBar::value = STRING

However, when I rewrite, some static variables are not initialized. Despite the fact that it just compiles.

Ported Code (not working)

#include <iostream>

struct String
{
    static const char * value;
};
const char * String::value = "STRING";

template<typename Base>
class Derived
{
public:
    static const char * value;
};
template<typename Base>
const char * Derived<Base>::value = { Base::value };

template<typename BarType>
struct Foo
{
    static const char * value;
};
template<typename BarType>
const char * Foo<BarType>::value = { BarType::value };

using Bar = Derived<String>;
using FooBar = Foo<Bar>;

int main()
{
    std::cout << "FooBar::value = " << FooBar::value << std::endl;
}

Fingerprints:

// nothing (Segmentation fault)

• Why is this happening?

• How do I fix / get around it?

This can be reproduced in Clang and Visual-C ++, however GCC FooBar::value = STRINGalso prints in the second example.

Update: Working Solution

@serge-ballesta. , . , constexpr VS.

Answer 1

, [basic.start.init]:

,

Derived<Base>::value Foo<BarType>::value , . . , , .. value s .

, . :

Derived<Base>::value ==> 0
Foo<BarType>::value ==> 0
Derived<Base>::value ==> Base::value
Foo<BarType>::value ==> BarType::value

:

Derived<Base>::value ==> 0
Foo<BarType>::value ==> 0
Foo<BarType>::value ==> BarType::value
Derived<Base>::value ==> Base::value

Derived<Base>::value, Foo<BarType>::value "STRING". , , 0. , , .

Answer 2

@ .

. String , String::value () .

:

• init Foo :

  ...
  template<typename BarType>
  struct Foo
  {
      static const char * value;
          static void init() {
              Foo::value = BarType::value;
          }
  };
  
  template<typename BarType>
  const char * Foo<BarType>::value;
  
  using Bar = Derived<String>;
  using FooBar = Foo<Bar>;
  
  int main()
  {
      FooBar::init();
      std::cout << "FooBar::value = " << FooBar::value << std::endl;
  }
  

• value Foo:

  ...
  template<typename BarType>
  struct Foo
  {
      static const char * value() {
          return BarType::value;;
      }
  };
  
  using Bar = Derived<String>;
  using FooBar = Foo<Bar>;
  
  int main()
  {
      std::cout << "FooBar::value = " << FooBar::value() << std::endl;
  }