CODEWITHSUNDEEP
clinker-errorsexternalstatic-linking
Meganathan
Meganathan

Reputation: 25

How can a static variable from other file accessed here on the C program below?

#include <stdio.h>
#include <stdlib.h>
int function(int);

int main()
{
    int a=1;
    extern b;//can be only declared
    printf("a=%d\n",function(a));
    printf("b=%d\n",value());
    return 0;
}
int function(int v)
{
    return v+1;
}

In file1.c :

static int b=20;

int value() //can't be used on a static function
{
    return b;
}

In the code above though I've made variable b as static, it's accessible from other files. How can it be accessed, as the static keyword shouldn't allow the variable to be accessed from other files right? I'm bit confused.

Upvotes: 1

Views: 780

Answers (2)

H.S.
H.S.

Reputation: 12679

This statement

static int b=20;

means the variable b will only be visible in its translation unit i.e. in file1.c, in which you have defined it.

In the main(), which is supposed to be in a different file, you have only declared b but not used it. Try to use it and you will find that linker will fail because it will not find the definition of b.

int main()
{
    int a=1;
    extern int b;    // note the type int in declaration. You have missed it.
    printf("access b = %d\n", b);   // <==== this won't work
    return 0;
}

Try to compile your program with above changes and you will see that the linker will throw error.

With above mentioned changes in main(), just remove the static keyword from variable b definition in file1.c file and above program will link successfully because then b will be a global variable which has external linkage.

From the value() you are returning the value that variable b hold which has nothing to do with whether b is static or non-static.

The lifetime of a static variable is the entire run of the program. You can access the static variable outside of it's scope/translation unit if you have the address of that static variable. Check this:

File file1.c :

#include <stdio.h>

static int b = 20;

int * retaddr_b (void) {
    return &b;
}

void print_b (void) {
    printf ("b = %d\n", b);
}

File main.c:

int * retaddr_b (void);
void print_b (void);

int main(void)
{
    print_b();                 // print static variable b value 
    int * ptr_b = retaddr_b(); // ptr_b pointing to b
    *ptr_b = 99;               // dereference ptr_b and assign different value
    print_b();                 // print static variable b value
    return 0;
}

Output:

b = 20
b = 99

Upvotes: 3

IrAM
IrAM

Reputation: 1738

Its not really accessed there, dint you see below warning when compiled:

warning: type defaults to ‘int’ in declaration of ‘b’ [-Wimplicit-int]

All you are doing is printing the return value from function value.

if you really wanted to see whether it is accessed or not check as below:

comment your printf("b = %d\n",value()); and use printf("b = %d\n",b);
in its place then compile and check.

Upvotes: 1

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