内存中的不同程序，相同变量，相同地址

Question

I have two C codes. test.c is

#include <stdlib.h>

int main ()
{
    int a;
    a = 5;
    return a;
}

test2.c is

#include <stdlib.h>

int main ()
{
    int a;
    a = 6;
    return a;
}

When I run them and I check the address in memory of the "a"s with gdb I get the same address. Why is this so?

Breakpoint 1, main () at test.c:7 7 return a; (gdb) print &a $1 = (int *) 0x7fffffffe1cc

Breakpoint 1, main () at test2.c:7 7 return a; (gdb) print &a $1 = (int *) 0x7fffffffe1cc

Answer 1

The address of "a" is on the stack frame for your program. This is a virtual address, independent of where in physical memory your program is actually loaded. Therefore, it would not be surprising if both (almost identical) programs used the same address.

Answer 2

Because each application in OS is run in its own memory space.

Address 0x7fffffffe1cc is not really physical address. This is made due to security - you cannot handle other process memory directly just like that. You also cannot handle devices directly.

You can read more about that here and here

Answer 3

It is very likely that your OS is using Virtual Memory for memory management. What this means is that addresses found within a given program are not 1:1 mapped to physical memory. This allows for a number of things (including running multiple programs that require lots of memory by page swapping to disk). Without virtual memory, if you were to allocate static int a rather than put it on the stack, the linker would do it's best to choose an address for it. If you then linked another program, it doesn't know what other programs may be using that address. Running two programs could stomp on the memory of the other program. With virtual memory, each program gets it's own slice of memory with it's own address 0x0 and it's own address 0x7fffffffe1cc .