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From: wallace@cme-durer.ARPA (Evan Wallace)
Newsgroups: comp.os.misc
Subject: Re: a very naive Question???
Summary: How can virtual memory be physical
Message-ID: <664@cme-durer.ARPA>
Date: 13 Oct 88 15:58:36 GMT
References: <835@amethyst.ma.arizona.edu> <5085@medusa.cs.purdue.edu>
Organization: National Bureau of Standards, Gaithersburg, MD
Lines: 29

In article <5085@medusa.cs.purdue.edu>, spaf@cs.purdue.edu (Gene Spafford) writes:
> The definition I use in teaching OS courses is the following:
> 
> If the address presented to the system may be bound to a physical
> address different from that seen by the program, you have
> virtual memory.  That is, if the software generates some address
> like 5 for a storage location, and if the binding to a physical
> address COULD POSSIBLY be to a hard physical address different
> from 5, then that implies virtual memory (let's not try to
> twist this to include fault handling, etc.).

This is mapped memory and not virtual, since the memory accessed is
indeed existant.

> Virtual memory does not mean that the virtual address range is bigger
> than the physical address range -- it could be smaller, as in

In "An Introduction to Operating Systems" by Deitel virtual storage
is defined to be a storage area larger than the primary storage
area (memory).  This is in direct contradiction with G. Spafford's
definition, but in complete agreement with common sense.  Also this
definition is even in my webster's dictionary with a 1966 date of
origin.?

> That's how I've taught it in OS courses at both Purdue and
> Georgia Tech, and that's consistent with many OS texts....

It's unfortunate that this kind of thing gets passed on to students
who will then continue its propagation.