Static variable inside a template base class

Question

This prints nothing:

#include <iostream>

template <typename Derived>
struct A
{
    static int test()
    {
        std::cout << "test" << std::endl;
        return 0;
    }

    static inline int a = test();
};

struct B : public A<B>
{
};

int main(int argc, char** argv)
{
    return EXIT_SUCCESS;
}

But this does:

#include <iostream>

template <typename Derived>
struct A
{
};

struct B : public A<B>
{
    static int test()
    {
        std::cout << "test" << std::endl;
        return 0;
    }

    static inline int a = test();
};

int main(int argc, char** argv)
{
    return EXIT_SUCCESS;
}

And also this:

#include <iostream>

struct A
{
    static int test()
    {
        std::cout << "test" << std::endl;
        return 0;
    }

    static inline int a = test();
};

struct B : public A
{

};

int main(int argc, char** argv)
{
    return EXIT_SUCCESS;
}

Not sure why or a workaround. I need the 'Derived' type to register it into a static table.

@user463035818 Presumably that you have to write it in every derived class. — Max Langhof, Oct 25 '18 at 15:03
@MaxLanghof not if `test` is put in a (non-templated) base clas — 463035818_is_not_an_ai, Oct 25 '18 at 15:16

score 2 · Answer 1 · answered Oct 25 '18 at 15:07

The reason why the first snippet doesn't print anything is that the static variable is not instantiated. You have to use that variable in order to instantiate it.

[temp.inst]/2

The implicit instantiation of a class template specialization causes the implicit instantiation of the declarations, but not of the definitions, default arguments, or noexcept-specifiers of the class member functions, member classes, scoped member enumerations, static data members, member templates, and friends

As a workaround, you can just use that variable:

int main(int argc, char** argv)
{
    (void) B::a;
    return EXIT_SUCCESS;
}

Yeah the implicit question is why the variable isn't initialized. The idea is to avoid any other code so the static function is executed for every derived class automatically. — chila, Oct 25 '18 at 15:52

score 1 · Answer 2 · answered Oct 25 '18 at 15:07

1

Since A is a template class, the static inline function/variable are not actually instantiated from the template unless they are used. Thus, you could do e.g. this:

#include <iostream>

template <typename Derived>
struct A
{
    static int test()
    {
        std::cout << "test" << std::endl;
        return 0;
    }

    static inline int a = test();
};

struct B : public A<B>
{
    static inline int b = a;
};

int main(int argc, char** argv)
{
    return 0;
}

Demo

answered Oct 25 '18 at 15:07

Max Langhof

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The idea is automation. – chila Oct 25 '18 at 15:54
1

@chila So why not use answers like [this](https://stackoverflow.com/questions/10332725/how-to-automatically-register-a-class-on-creation)? – Max Langhof Oct 25 '18 at 16:01
The registration should occur at program start for all the classes that inherit from the templated class, before main is executed. – chila Oct 25 '18 at 16:04
I want to register the class type in a static set. – chila Oct 25 '18 at 16:06
1

That sounds exactly like the answer I linked... Did you read it? – Max Langhof Oct 25 '18 at 16:12

chila · Answer 3 · 2018-10-25T17:25:47.097

The (automatic) solution, as pointed here is to create a constructor that uses the registration variable. Also, the variable will be initialized only if an object is constructed.

#include <iostream>

template <typename Derived>
struct A
{
    A()
    {
        a = 0;
    }

    static int test()
    {
        std::cout << "test" << std::endl;
        return 0;
    }

    static inline int a = test();
};

struct B : public A<B>
{
};

int main(int argc, char** argv)
{
    B b;

    return EXIT_SUCCESS;
}

The 'a = 0' is to avoid a warning of unused variable. Overhead should be minimal.

Demo

Static variable inside a template base class

3 Answers3