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From: swarren@convex.com (Steve Warren)
Newsgroups: comp.sys.amiga
Subject: Re: A2630 board
Message-ID: <100428@convex.convex.com>
Date: 7 Mar 90 16:06:34 GMT
References: <1745@crash.cts.com>
Sender: news@convex.com
Followup-To: comp.sys.amiga.hardware
Organization: Convex Computer Corporation; Richardson, TX
Lines: 58

In article <1745@crash.cts.com> hawk@pnet01.cts.com (John Anderson) writes:
                                [...]
>"thing" needs 32 bits then that means it can only hold 512 thousand "Things"
>since each one is 32 bits long?  Is that correct?  And if so, or not so, how
>much RAM will be left over after your SetCPU program copies Kickstart into 32
>bit RAM?  
                                [...]
The "Things" you are talking about are called words, and the definition of
word-size has nothing to do with how much capacity a memory has, only how
it is organized.

Here is how 16 bytes of data would be stored in 16-bit ram (labeled A-P):

ADDRESS		DATA (2 bytes/16-bit word)

00000000	B,A
00000002	D,C
00000004	F,E
00000006	H,G
00000008	J,I
0000000A	L,K
0000000C	N,M
0000000E	P,O

(The addresses are in hexadecimal; byte 'A' is stored at '0', 'B' at '1', etc)


Here is how the same 16 bytes of data would be stored in 32-bit ram:

ADDRESS		DATA (4 bytes/32-bit word)

00000000	D,C,B,A
00000004	H,G,F,E
00000008	L,K,J,I
0000000C	P,O,N,M


Notice that sequential word-addresses in the 16-bit memory are incremented
by two, but sequential word-addresses in the 32-bit memory are incremented
by four.  The address actually corresponds to a single byte location (the
byte farthest to the right, in this example).

So each word is twice as long, and it only takes half as many words to
store the same program.  Since the addresses of each byte are unchanged
(the 680X0 family maintains an address for each byte in memory, regardless
of bus-size), the storage capacity of the memory is unchanged.  But the
68030 can access the data in 32-bit chunks and then sort it out, so it
goes faster.

In other words, if you use the same number of memory chips - ie the byte-count
is the same, you get the same total storage capacity, regardless of how the
memory is organized.  Addresses are not related to word-size either.  But
the 32-bit storage will work much faster.

--
--Steve
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