Operating system – Why is the address of the variable remains the same in the modified Fork () system call

if (fork() == 0)
{
 a = a + 5;
 printf("%d, %d 
", a, &a);
}
else
{
 a = a-5;
 printf ("%d, %d 
", a,& a);
}

Please consider the following code snippet.

if (fork() == 0)
{
 a = a + 5;
 printf("%d, %d 
", a, &a);
}
else
{
 a = a-5;
 printf ("%d, %d 
", a,& a);
}

AFAIK, when creating fork(), the parent The virtual address space is copied to the child, and the child & parent share the same physical page until one of them tries to modify. At the moment one of the child & parent modifies the variable, paren The physical page of t is copied to another page of the child page, and the physical page remains private.
So, the value of’a’ here is different in child&parent. But when it comes to children and children’s’a’ When the address is the parent, the output is the same. Even if the physical page is different, I can’t figure out why the address remains the same.

The address of a is not the actual physical address .

This is a virtual address.
The hardware/operating system layer maps the virtual address to a physical address (not visible to your application).

Therefore , Even if the addresses have the same number, they will not be mapped to the same physical memory on the ram chip.

PS. It is better to use “%p” when printing addresses (ie pointers) with printf()< /p>