No linkage at block scope?

Question

Do all variables declared in a block have \'no linkage\'?

For example:

1:

If I declare a static variable:

void foo()
{
   static int i;         


        

Answer 1

1. "Would it have internal linkage or no linkage? if no linkage then why make it static?" -- it would have no linkage. static specifies the static storage duration.

2. "What happens if i use extern?" It will be a declaration of a name with external linkage, and since there is none in global scope, the program will report linkage errors. Edit: Since there is a previous static declaration visible in the scope, the standard says the name "receives the linkage of the previous declaration" 3.5/6, so the i inside foo() will have internal linkage.

Answer 2

Indeed, 'no linkage' at function scope.

The goal is lifetime management: the static has the lifetime of a global static, while it has the visibility (scope) of a local.

Note

In C++ you can also declare statics ('globals') without linkage by enclosing them inside an anonymous namespace. This trick is used commonly in header-only libraries:

namespace /*anon*/
{
    void foo() {}    // only in this translation unit
    int answer = 42; // this too
}

What happens if I use extern?

If you use extern, the declaration is an extern declaration only (nothing is defined). As such, it normally would be expected to external linkage by definition - being defined in another translation unit. (So it acts the same as if when it was declared at global scope). This is similar to local function declarations:

int main()
{
    void exit(int); // equivalent to non-local declaration
}

Note that, in your 2. example, variable i was already declared static and it will therefore not get external linkage. I might get declared in another translation unit without linker conflicts, though.