Hello...
> Process A has a memory allocated and memory pointer x pointing to
> that memory. I fork process B from a running process A.B gets a new
> copy of the memory and the pointer.However, printing x in A and B
> points to the same memory location(though actually differnt physical
> memory locations).What is the concept of VM here.Can anyone kindly
> explain ?
First, strace is your friend here :) use it to find out how fork() is
eventually transformed into clone(). You will watch that there is no
CLONE_VM there, so child is actually given a separate address
space...but initially they are pointing to the same pages. Only when
one of them is writing something to these pages (stack, heap, etc),
Copy on Write is triggered, thus the writer is assigned unique pages.
In your case, actually they are pointing to different location of
physical memory. But remember, we "live" under the mercy of virtual
memory manager. Even though pointer a points to different physical
address, virtual memory manager is able to manipulate the page table
thus it points to identical virtual memory address.
More specifically, a is variable which lives in stack area. Stack area,
especially in non randomized address space enabled-kernel, always start
more or less in same virtual address. So, even you fork a child, kernel
tends to assign same virtual memory address for stack pointer. When a
local variable is allocated, stack pointer is decremented (in
architecture where stack grows to the lower address) with the same
number, there's why you got same output on both child and parent.
Observe the assembly instruction of your fork test program and you will
get better understanding. Usually, this code block:
<snip>
void main()
{
char a=0;
<snip>
will be transformed into this:
804835c: 55 push %ebp
804835d: 89 e5 mov %esp,%ebp
804835f: 83 ec 08 sub $0x8,%esp --> allocating "a"
8048362: 83 e4 f0 and $0xfffffff0,%esp --> aligning "a"
8048365: b8 00 00 00 00 mov $0x0,%eax
804836a: 29 c4 sub %eax,%esp
Got the picture now? If stack VMA starts from same virtual address, "a"
as result of decrementing %sp, will end in same virtual address too.
Pheww, this is a bit tricky for me. I hope I can explain in very clearly
in one shot :)
regards,
Mulyadi
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