Static variable in an inline method across compilation units - c++

Question

Consider the following header file Sample.h:
#pragma once
template<typename T> class Sample {
static T Method() {
static T var = T(0);
var++;
return var;
}
};
int U1Test();
int U2Test();
And 2 compilation units, U1.cpp:
#include "Sample.h"
int U1Test() { return Sample<int>::Method(); }
And U2.cpp:
#include "Sample.h"
int U2Test() { return Sample<int>::Method(); }
Then in another unit Main.cpp:
#include "Sample.h"
#include <iostream>
using namespace std;
int main() {
cout << U1Test() << endl;
cout << U2Test() << endl;
return 0;
}
When compiled, it gives me the following output:
1
2
But I'm not sure how the compiler does this, because the method is inline header-only and there is no compilation unit for it. So I would expect each compilation unit (like U1.cpp and U2.cpp) to receive its own copy of var because the method is inlined in that compilation unit.
Is there a subtle change that would make the variable separate in each compilation unit? I'm asking because code like this in a larger program seems to lead to crashes, so perhaps my reproducer is not enough (the reproducer works according to the C++ standard, AFAIK).
The compiler is g++ (conda-forge gcc 10.3.0-16) 10.3.0 on Ubuntu 20.04.

Answer 1

So I would expect each compilation unit (like U1.cpp and U2.cpp) to receive its own copy of var because the method is inlined in that compilation unit.
(The implicit) inline means that there may be multiple copies of the same function, but the linker will ignore all but one of them. Thus, all compilation units access that one function instantiation.
Is there a subtle change that would make the variable separate in each compilation unit?
Yes, make the function (freestanding) static rather than (implicitly) inline:
static T Method() {
static T var = T(0);
var++;
return var;
}
Note that this function is not in a class.