Unique address for constexpr variable

If you need to take the address of constexpr variable, declare it as a static member variable. It can be used as a constant expression this way (as opposed to using a function returning a const).

foo.h:

#ifndef FOO_H
#define FOO_H

struct Foo {
  static constexpr int foo { 42 }; // declaration
};

#endif // FOO_H

foo.cpp:

#include "foo.hpp"

constexpr int Foo::foo; // definition

bar.cpp:

#include "foo.hpp"

const int* foo_addr() {
  return &Foo::foo;
}

int foo_val() {
  return Foo::foo;
}

main.cpp:

#include <iostream>
#include "foo.hpp"

extern const int* foo_addr();
extern int foo_val();

constexpr int arr[Foo::foo] {}; // foo used as constant expression

int main() {
  std::cout << foo_addr() << " = " << foo_val() << std::endl;
  std::cout << &Foo::foo << " = " << Foo::foo << std::endl;
}

Output:

$ g++ -std=c++11 foo.cpp bar.cpp main.cpp -o test && ./test
0x400a44 = 42
0x400a44 = 42

C++17 inline variables

This awesome C++17 feature allow us to:

• conveniently use just a single memory address for each constant
• store it as a constexpr: How to declare constexpr extern?
• do it in a single line from one header

main.cpp

#include <cassert>

#include "notmain.hpp"

int main() {
    // Both files see the same memory address.
    assert(&notmain_i == notmain_func());
    assert(notmain_i == 42);
}

notmain.hpp

#ifndef NOTMAIN_HPP
#define NOTMAIN_HPP

inline constexpr int notmain_i = 42;

const int* notmain_func();

#endif

notmain.cpp

#include "notmain.hpp"

const int* notmain_func() {
    return &notmain_i;
}

Compile and run:

Compile and run:

g++ -c -o notmain.o -std=c++17 -Wall -Wextra -pedantic notmain.cpp
g++ -c -o main.o -std=c++17 -Wall -Wextra -pedantic main.cpp
g++ -o main -std=c++17 -Wall -Wextra -pedantic main.o notmain.o
./main

GitHub upstream. See also: How do inline variables work?

C++ standard on inline variables

The C++ standard guarantees that the addresses will be the same. C++17 N4659 standard draft 10.1.6 "The inline specifier":

6 An inline function or variable with external linkage shall have the same address in all translation units.

cppreference https://en.cppreference.com/w/cpp/language/inline explains that if static is not given, then it has external linkage.

Inline variable implementation

We can observe how it is implemented with:

nm main.o notmain.o

which contains:

main.o:
                 U _GLOBAL_OFFSET_TABLE_
                 U _Z12notmain_funcv
0000000000000028 r _ZZ4mainE19__PRETTY_FUNCTION__
                 U __assert_fail
0000000000000000 T main
0000000000000000 u notmain_i

notmain.o:
0000000000000000 T _Z12notmain_funcv
0000000000000000 u notmain_i

and man nm says about u:

"u" The symbol is a unique global symbol. This is a GNU extension to the standard set of ELF symbol bindings. For such a symbol the dynamic linker will make sure that in the entire process there is just one symbol with this name and type in use.

so we see that there is a dedicated ELF extension for this.

I think constexpr is meant more for functions whose return value is constant. You can bind a constant variable to the return value of a constexpr function and expose that externally instead. For example:

// constexpr.h
#ifndef __CONSTEXPR_H
#define __CONSTEXPR_H

extern const int foo;

#endif // __CONSTEXPR_H

// constexpr.cpp
#include "constexpr.h"

constexpr int foo_expr()
{
    return 42;
}

const int foo = foo_expr();

// unit1.cpp
#include <iostream>
#include "constexpr.h"

void unit1_print_foo()
{
    std::cout << &foo << " = " << foo << std::endl;
}

// unit2.cpp
#include <iostream>
#include "constexpr.h"

void unit2_print_foo()
{
    std::cout << &foo << " = " << foo << std::endl;
}

// main.cpp
extern void unit1_print_foo();
extern void unit2_print_foo();

int main(int, char**)
{
    unit1_print_foo();
    unit2_print_foo();
}

My result is:

$ g++-4.7 -std=c++11 constexpr.cpp unit1.cpp unit2.cpp main.cpp -o test && ./test
0x400ae4 = 42
0x400ae4 = 42

However, it should usually be sufficient to make the foo_expr function itself externally visible, and callers would use foo_expr() to get the value instead of treating it like a variable.