Re: NegativeArraySizeException ... IndexOutOfBoundsException ...

=?ISO-8859-1?Q?Arne_Vajh=F8j?= <>
Sun, 17 Jan 2010 19:58:40 -0500
On 12-01-2010 14:08, Lew wrote:

On Jan 12, 11:16 am, Wojtek<> wrote:

Arne Vajh?j wrote :

On 11-01-2010 17:09, Maarten Bodewes wrote:

Arne Vajh?j wrote:

BTW, even long would be too small for indexes if Java
will be used after 2074, but somehow I doubt that would
be the case. And besides we do not have the verylong datatype

You are expecting memory sizes of 9,223,372,036,854,775,807 bytes????
That's 9,223 PETA bytes. Hmm, weird, may happen. But it is certainly
rather large.

In 2074 ? Yes !

I am curious. Just what WOULD need such a large index? Every sand grain
on earth? Every star in every galaxy in the universe?

Where will you store the array? Either you have a crapload of RAM
(one bit per atom storage density?) or the largest-capacity storage
device ever invented (one bit per atom storage density?).

What's the average retrieval latency? Even at one bit per atom
storage density, it must take even a light beam noticeable time to
reach the further reaches of the storage device; anything slower like
a semiconductor must take a really long time.

A silicon crystal lattice has a lattice spacing of just over half a
nanometer, or 5.4 x 10^-10 m. A three-dimensional storage medium for
a 9 x 10^18-element array would hold juar over 2 x 10^6 elements to
the side. An average access would be halfway in each dimension, or
10^6 elements, which in a silicon lattice is about 5.4 x 10-4 m, times
three for a total traversal distance of about 1.6 x 10^-3 m. Each
way. For a round trip slightly over 3 x 10^-3 m. A light beam
travels that in 0.1 microseconds (10^-7 s). That's about 200 clock
cycles of latency on a modern processor, far more on the future
processors of 2074.

With correction:
 >Drat! Mixed up my CGS and MKS. That's 10-5 s, or 10 microseconds.

Neither looks correct to me.

s = 3 x 10^-3 m
v = 3 x 10^8 m/s
t = s/v = 1 x 10^-11 s


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