why function didnt need extern, but variable does

Question

Sorry guys I know my english is bad, but i made examples so that my question is more clearer.

a.cpp

#include <iostream>

using namespace std;

void funcfoo(){
    cout << "test only" << endl;
}

int varfoo = 10;

b.cpp

#include <iostream>

using namespace std;

extern void funcfoo();

extern int varfoo;

int main(){

    funcfoo();

    cout << varfoo;

    return 0;
}

Then I compile it like this "cl b.cpp a.cpp"

My question is. How come when I remove the "extern keyword before void funcfoo()" it works fine, but when i remove the extern keyword before int var foo I get an error?

Answer 1

The issue is what each one of the lines of code means. int varfoo is a definition of a variable, while void funcfoo() is only a declaration. You can provide multiple declarations of an entity, but only one definition. The syntax to provide a declaration and only a declaration of a variable is by adding the extern keyword: extern int varfoo; is a declaration

---

3.1 [basic.def]/2 A declaration is a definition unless it declares a function without specifying the function’s body (8.4), it contains the extern specifier (7.1.1) or a linkage-specification25 (7.5) and neither an initializer nor a function body [...]