Thread: Linkage vs scope of a variable

What is the contradiction?

Originally Posted by MartinR

Well I think it is obvious - How can it be accessed by other file if it has file scope in the same time?

Consider this program consisting of two source files:

Code:

/* main.c */

#include <stdio.h>

void foo(void);

void bar(void)
{
    extern int i;
    printf("i=%d\n", i);
}

void baz(void)
{
    printf("i is equal to %d\n", i);
}

int main(void)
{
    foo();
    bar();
    baz();
}

Code:

/* foo.c */

int i;

void foo(void)
{
    i = 123;
}

If you compiled the above program, you would find that the compiler calls you out on an error in main.c concerning the baz function, i.e., it would complain about i not being declared or something like that.

The reason is that the name i has file scope in foo.c, but in main.c its scope is that of the bar function, hence it is out of scope in the baz function. Yet, because i has external linkage, the i that was declared extern in bar refers to the same int object as the i that was declared at file scope in foo.c