How do c++ compilers find an extern variable?

Question

I compile this program by g++ and clang++. There has a difference:
g++ prints 1, but clang++ prints 2.
It seems that
g++: the extern varible is defined in the shorte

Answer 1

Clang gives a right result. Even though per the rule of the current standard, the program shouldn't be ill-formed. Note the emphasized wording:

If, within a translation unit, the same entity is declared with both internal and external linkage, the program is ill-formed.

The entity declared at #3 and the entity declared at #1 are not the same entity, because of the below rule:

Two names that are the same and that are declared in different scopes shall denote the same variable, function, type, template or namespace if

• both names have external linkage or else both names have internal linkage and are declared in the same translation unit; and
• [...]

They're not, one has internal linkage and the other has external linkage, so they do not denote the same entity, hence the code does not violate the [basic.link#6]. Moreover, the example follows [basic.link#6] remains a wrong interpretation about variable i.

P1787 has clarified this example. It says:

static void f();
extern "C" void h();
static int i = 0;               // #1
void gq() {
  extern void f();              // internal linkage
  extern void g();              // ::g, external linkage
  extern void h();              // C language linkage
  int i;                        // #2: i has no linkage
  {
    extern void f();            // internal linkage
    extern int i;               // #3: exinternal linkage
  }
}

Without Even though the declaration at line #2 hides the declaration at line #1, the declaration at line #3 would link with thestill redeclarationes at line #1. Because the declaration with and receives internal linkage is hidden, however, #3 is given external linkage, making the program ill-formed.

That means, In your example, the variable i introduced by declaration extern int i will link with the variable i declared by static int i. So, print 2 is the right behavior.