How to declare constexpr extern?

Question

Is it possible to declare a variable extern constexpr and define it in another file?

I tried it but the compiler gives error:

De

Answer 1

Yes it somewhat is...

//===================================================================
// afile.h

#ifndef AFILE
#define AFILE

#include 
#include 

enum class IDs {

  id1,
  id2,
  id3,
  END

};

// This is the extern declaration of a **constexpr**, use simply **const**
extern const int ids[std::size_t(IDs::END)];

// These functions will demonstrate its usage

template void Foo() { std::cout << "I am " << id << std::endl; }

extern void Bar();

#endif // AFILE

//===================================================================
// afile.cpp

#include "afile.h"

// Here we define the consexpr. 
// It is **constexpr** in this unit and **const** in all other units
constexpr int ids[std::size_t(IDs::END)] = {

  int(IDs::id1),
  int(IDs::id2),
  int(IDs::id3)

};

// The Bar function demonstrates that ids is really constexpr
void Bar() {

  Foo();
  Foo();
  Foo();

}

//===================================================================
// bfile.h

#ifndef BFILE
#define BFILE

// These functions will demonstrate usage of constexpr ids in an extern unit

extern void Baz();
extern void Qux();


#endif // BFILE

//===================================================================
// bfile.cpp

#include "afile.h"

// Baz demonstrates that ids is (or works as) an extern field
void Baz() {

  for (int i: ids) std::cout << i << ", ";
  std::cout << std::endl;

}

// Qux demonstrates that extern ids cannot work as constexpr, though
void Qux() {

#if 0 // changing me to non-0 gives you a compile-time error...

  Foo();

#endif

  std::cout << "Qux: 'I don't see ids as consexpr, indeed.'" 
            << std::endl;

}

//===================================================================
// main.cpp

#include "afile.h"
#include "bfile.h"

int main(int , char **)
{

  Bar();
  Baz();
  Qux();

  return 0;
}