Re: A couple of DRAM memory stick questions ??

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On Wed, 2009-09-30 at 10:55 -0400, William Case wrote:
> I'll take that information to the bank.  To state it another way just
> to make sure I've got it.  A typical physical address goes to, or
> points to,  8 + 8 + 8 + 8 + 8 + 8 + 8 + 8 cells arranged side-by-side
> in a line on an individual DIMM/DRAM stick. 

+8 more if it's parity memory, or more for ECC memory.

> I suspect that by thinking of address as divided into bytes rather
> than a single 64 bit word (dword, qword, -- pick your author) there is
> a natural division for instructions, numbers and characters within the
> 'word'. Or, is there some physical reason why it is thought of as 8 +
> 8 ...

It's the data that's stored in units of 8 bits. When addresses are
stored then of course the same applies. When they're on the address
lines of the memory bus, they may be in groups of 16 or 32 or 64
(depends on the bus design). None of this matters to you as a

Note that the pedantic name for a group of 8 bits is "octet". A "byte"
is the number of bits required to represent a character in some
encoding. Nearly all modern machines have 8-bit bytes and are
byte-addressed, so we tend to equate "byte" with "8 bits", but I've used
machines where addresses refer to 12-bit words, and the DEC-10 famously
had 36-bit words and a configurable byte size, usually set to 6.

> When you say "chips" above I assume you mean cell, i.e. chip = cell =
> 1 capacitor and 1 transistor for storage of 1 bit.

A chip has a whole bunch of cells (in the millions these days). They
aren't the same.


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