What happens if you try to compile and run this program?

What happens if you try to compile and run this program?

#include <stdio.h>

int *f();

int main (int argc, char *argv[]) {

int *p;

p = f();

printf("%d",*p);

return 0;

}

int *f() {

static v = 1;

return &v;

}

Choose the right answer:
A . The program outputs 1
B . Compilation fails
C . The program outputs 3
D . The program outputs 2
E . The program outputs 0

View Answer

Answer: A

Explanation:

The program outputs 1 because the static variable v is initialized to 1 inside the f function, and it is visible to the main function. The f function returns the address of v, which is a pointer to an int. The main function dereferences the pointer and assigns it to p, which is another pointer to an int. Then, the main function prints the value of *p, which is the same as dereferencing p again. Therefore, the output of the program is:

f() = &v p = f() printf(“%d”,*p) = &v = 1

The other options are incorrect because they either do not match the output of the program or do not use the correct concept of static variables.