Re: The Linear Address Space
From: Cameron Gibbs (odly_at_hotmail.com)
Date: 04/29/04
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Date: Thu, 29 Apr 2004 06:07:14 +0000 (UTC)
I think I am getting there.
Please correct where necessary.
Let's forget the distinction between the legacy I/O address space for now.
The Address Space *is* all of the addresses that a processor can use!
When a computer starts up, hardware sets up the addresses for things like
the Hard Drive which works to assign address values to things like it's
accessible registers.
This would probably also mean that all of the processor registers, the
video card registers etc. would all be given addresses.
This is probably what is meant by the term "mapping."
A programmer uses these registers (at their assigned address) to write
binary numbers to it that act to command the device.
In the case of a Hard Drive, the registers can be used to send packets of
data (512b or 1k packets) to RAM which are then executed or read by a
program.
So if you want to access a byte of Hard Disk space, one first commands it's
sector to read it into RAM and then selects it's offset.
A few things still to clear up:
If for example a program has a loop in it, then at some stage it has to
point (address) back to a previous part of itself.
Obviously if it is executing it is in RAM and hence doesn't use the Hard
Drive sector address and offset so what address does it use?
Does RAM have it's own set of addresses?
Do Operating Systems have to manage the loop link addresses if more than one
program is in RAM or are they in some way taken care of when programs are
put into RAM?
Also:
Paging handles the virtual RAM feature which manages the swap space to
increase the effective size of RAM by using Hard Disk space.
So paging is not actually necessary for the execution of a program and RAM
is in fact used regardless of paging.
Thanks to all who are helping.
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- Previous message: Matt Taylor: "Re: debug registers"
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