Re: Linux VM Question

["Fawad Lateef" <fawadlateef@xxxxxxxxx>]

On 9/4/06, pandari kashyap <pandari.kashyap@xxxxxxxxx> wrote:
> Thank you for your reply. So are you telling me that user mode processes (A
> and B) have the same set of possible linear address spaces ( 0 to 4GB). For
> example more realistically A has address space beginning at address 24 and
> ending at address 124 and B also has address space beginning at 24 and
> ending at 124.

Yes, this can be the case

> Let us say that A and B are 2 completely different processes. When either A
> or B accesses, for example, address 30 then how does kernel resolve it to
> the correct physical address?.
>
> Because all the kernel has, at this point in time, is linear address 30 and
> nothing else? Can you explain more? I appreciate your help

The linear/virtual address translation to physical address is done
through page-tables and each process has its own page-table. So at
context-switching time the process's (which is going to execute)
page-table base address is loaded and then the linear/virtual address
points to a new physical address currently in use by process.

For getting linear/virtual to physical address translation you can
consult Intel's x86 System Programming Manual or read a good operating
systems book.

--
Fawad Lateef

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