Subhobroto Sinha wrote:
....
> Oh, and a quickie :
>
> When we print an address of a variable, is it the
> virtual address or the physical one which is printed
?
....
> Look at it closely, and match it with the output.
Who
> can explain that to me ?
I am no expert on Linux memory management but even so
I feel there can be no question about it: what is
printed is the virtual address.
I have no definite explanation about why the addresses
of i in the parent and child processes appear to be
the same. (They certainly _are not_ the same, since
modifying i in the child does not modify i in the
parent.) I can only offer a guess. But it might be
useful first to read this snippet from the vfork
manpage:
<snippet>
Under Linux, fork() is implemented using
copy-on-write pages, so the only penalty incurred by
fork() is the time and memory required to duplicate
the parent's page tables, and to create a unique task
structure for the child. However, in the bad old days
a fork() would require making a complete copy of the
caller's data space, often needlessly, since
usually immediately afterwards an exec() is done.
</snippet>
As it says here, the entire data space of the parent
is not duplicated for the child; instead only the page
tables are copied. And, therefore, my guess is that
the address of i (or any other variable) as printed is
an _offset_ from some base address which we don't
know. The base address is different for the parent and
the child. This is consistent with the philosophy of
saving time: merely change the base address in the new
page tables and you have a complete new address space.
If you find an authentic explanation please share it
with me.
- Manas Laha
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