Why in a C++ template, declaring a vector member variable as static will produce linker error but inline static will not?

Question

I cannot explain the following (using C++20 rules)

//FooTemplate.h
template<typename T>
class FooTemplate {
private:
   static std::vector<T> t; //linker error.
   //inline static std::vector<T> t; --> will not produce linker error
   //static T t; -->this works also with the appropriate changes in the following code
public:
    static void setT(T t1)
    {
        t.push_back(t1);
    }
    static T getT(int i)
    {
        return t.at(i);
    }
};

//main.cpp
#include"FooTemplate.h"
int main() {
  FooTemplate<Foo>::setT(Foo{});
  FooTemplate<Foo>::getT(0);
}

I have to inline the static member t in order for it to work. If I do not use a templated container (e..g define T t instead of vector<T> t it also works.

Any idea?

I'm guessing you didn't try using the `static T t;` - Because that doesn't work. — StoryTeller - Unslander Monica, Jan 26 '21 at 07:27
On the contrary `static T t;` works (it does not produce linker errors) — user2465039, Jan 26 '21 at 07:36
Again, you likely didn't **use** that variable. I would bet `appropriate changes in the following code` involved simply commenting it out. Try taking it's address `(void)&t;` instead. — StoryTeller - Unslander Monica, Jan 26 '21 at 08:33

score 1 · Accepted Answer · answered Jan 26 '21 at 07:29

Static members must be defined (initialized) outside their class.

Simply write in your implementation:

template<class T> std::vector<T> FooTemplate<T>::t = [some initializer];

EDIT: You do not actually need to explicitely initialize them (unless they are objects of a class without a default constructor), but you do need to repeat their declaration like this anyway.

Why in a C++ template, declaring a vector member variable as static will produce linker error but inline static will not?

1 Answers1