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

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.