static variable result

Code:

---

#include<stdio.h>

void foo ()
{
        static int i = 0;
            printf("i = %d \n", i);
            i = 1;
                printf("i = %d \n", i);       
            i = 2;
                printf("i = %d \n", i);       
                i = 3;
                printf("i = %d \n", i);       
}
int main ()
{
        foo ();
        printf("\n");
        foo ();
        return 0;
}

---

First Call
i = 0
i = 1
i = 2
i = 3

Second Call
i = 3
i = 1
i = 2
i = 3

because of static keyword i become equal to 3. exactly. When code run what happen in program ?

Just remember that a static local variable retains its value after control returns to the caller. That's why the apparently initial value of i was 3 on the second call to foo.

Does this confuse you less?

Code:

---

#include<stdio.h>

static int i = 0;

void foo ()
{
            printf("i = %d \n", i);
            i = 1;
                printf("i = %d \n", i);       
            i = 2;
                printf("i = %d \n", i);       
                i = 3;
                printf("i = %d \n", i);       
}
int main ()
{
        foo ();
        printf("\n");
        foo ();
        return 0;
}

---

The only difference here is that the variable called i now has a scope that extends to the whole file, and not just the foo() function.