On Sun, Jan 11, 2009 at 4:44 PM, julius <jul...@juliuspaintings.co.uk> wrote:
> About This Mac says that I have 2GB of internal memory.
> Is this 2GB of 64-bit words or 2GB of 8-bit bytes?
> I appreciate that GB is Giga Byte but ......
Others have covered this adequately but I just want to reinforce that
there's essentially no other way to interpret 2GB these days other
than as roughly 16 billion bits. ("Roughly" because of the whole
decimal/binary silliness.)
> Similarly with respect to the L2 Cache, I have 12 MB per processor, is that
> 12 MB by 8 bits or 64 bits?
Likewise here. This principle applies *everywhere*.
> In thinking about memory usage, where previously I would think of my program
> in terms of 8 or 16 or 32 bit words should I now be thinking in terms of 64
> bit words?
> That is, should I think of my available internal memory space as effectively
> being 500MB words?
No, this makes no sense. You have 2GB of memory. If you're working
with 64-bit words then it makes sense to think of your memory as being
roughly 256 million words (note: not 500) but that's not the same as
"500MB words".
> Similarly, say that I had 100MB of 2 x 8-bit byte integers to save to disk,
> should I now think that this will be saved as 100MB by 64 bit (i.e. 8 x
> 8-bit byte) integers?
> If it is 100MB by 64 bit integers then should I think of compressing the
> data so as to reduce bandwidth requirements?
Just because your machine has a 64-bit processor doesn't mean you're
suddenly required to work with 64-bit quantities everywhere. You can
still work with 8, 16, or 32-bit quantities as you need. Which one is
the most appropriate choice, I couldn't say, but you seem to have this
strange idea that your 8-bit integers will somehow magically take up
64 bits of storage just because you're running on a Core2
architecture. It's simply not the case.
Mike
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