If a static variable is declared out side of a function, will the memory address be the same as if it's declared in a function

Question

I was asked these 2 questions during an interview. I guess the address of a static variable will be the same no matter where it is declared. It will also have the same address from run to run. Correct me if I am wrong.

1. If a static variable is declared out side of a function, will the memory address be the same as if it's declared in a function?

2. If a static variable is declared out side of a function, will it have the same address every time you run the program?

Answer 1

Cunningham's Law will ensure quick correction of my mistakes:

1. global and local static variables are treated similarly so they may indeed end up at the same address when moved:

#include <stdio.h>

#ifdef GLOBAL
static int i;
#endif
// static int j;

#ifndef GLOBAL
void f(void) {
    static int i;
    printf("local:  %p\n" , (void *) &i);
}
#endif

int main() {
#ifdef GLOBAL
    printf("global: %p\n" , (void *) &i);
#else
    f();
#endif
}

You need to disable address randomization to observe it:

$ sudo bash -c 'echo 0 >  /proc/sys/kernel/randomize_va_space';\
gcc 1.c && ./a.out &&\
gcc -DGLOBAL 1.c && ./a.out;\
sudo bash -c 'echo 2 >  /proc/sys/kernel/randomize_va_space'
local:  0x555555558034
global: 0x555555558034

If you add another global static int j (commented out above) after the global i then the effect goes away:

local:  0x555555558038
global: 0x555555558034

So call me maybe!?

2. No. Modern operating systems, specifically the program loader, use address space layout randomization (ASLR) so the address changes on every execution:

$ ./a.out && ./a.out
global: 0x563bcce14034
global: 0x55cd4a497034