Address of static data member

Question

Why C++ doesn't allow taking the address of a static data member when the data member is initialize within the class and doesn't have an out-of-class definition? How is the storage allocated for static member in this case?

The below minimum program demonstrate the issue.

#include <iostream>

class Test {
public:
    static const int a = 99;   // how is the storage allocated for Test::a??
};

// const int Test::a;

int main() {
    std::cout << Test::a << '\n';  // OK, print 99
    const int* ptr = &Test::a;     // Linker error, undefined reference to Test::a
}

If I uncomment the line const int Test::a, then the program works fine.

related/dupe: https://stackoverflow.com/questions/19630570/what-does-it-mean-to-odr-use-something — NathanOliver, Nov 03 '22 at 20:06
"_how is the storage allocated for Test::a??_": It doesn't allocate any storage, which is why you can't take the address and why a definition (which will instruct the compiler to allocate storage) is necessary. — user17732522, Nov 03 '22 at 20:06
the storage is effectively allocated at compile time - exactly how and where is an implementation detail — Neil Butterworth, Nov 03 '22 at 20:07
"If I uncomment the line "const int Test::a", then the program works fine" - There's no `const int Test::a` in the code you show. — Jesper Juhl, Nov 03 '22 at 20:07
If you declare the variable `a` inline everything works fine: `inline static const int a = 99` — Robert Shepherd, Nov 03 '22 at 20:11

Nelfeal · Accepted Answer · 2022-11-03T20:23:49.973

This line is a declaration:

static const int a = 99;

It is not a definition. For storage to be allocated, you need a definition. That's where your commented line comes into play.

You can use Test::a even if it has no definition because it is treated as a compile-time constant. In general, a static const T variable (where T is a trivial type) must be usable in a constant expression as soon as it is constant-initialized.

Note that you can get both behaviors (compile-time constant and variable with an address) at the same time:

class Test {
public:
    static const int a = 99;
};

const int Test::a;

template<int i>
struct Templated {};

int main() {
    const int* ptr = &Test::a;
    Templated<Test::a> a;
}

Demo

And to illustrate the constant-initialization point:

class Test {
public:
    static const int a;
};

template<int i>
struct Templated {};

// does not compile because Test::a is *not yet* usable in a constant expression
Templated<Test::a> global;

const int Test::a = 99;

int main() {
    const int* ptr = &Test::a;
    Templated<Test::a> a;
}

Demo

You could also just switch the member to `static constexpr int a = 99;` and now no out of line definition is required. — NathanOliver, Nov 03 '22 at 20:22

Address of static data member

1 Answers1