Re: spinoza programming language and big numbers

Zara said:

On Tue, 18 Mar 2008 00:48:49 +0100 (CET), Three Headed Monkey
<four_headed_monkey@xxxxxxxxx> wrote:


I am researching big numbers and read in this group that
spinoza programming language have unlimited integers.

I am trying to calculate this formula:

x = 9^(8^(7^(6^(5^(4^(3^(2^1)))))))

("^" is exponentiation operator)

People tell me that general purpose languages are not very good
with very big numbers, but Mr. Spinoza says spinoza programming
language will have unlimited numbers.

Can problem be expressed in spinoza programming language?

Where can I buy / download spinoza compiler?


Why don't you try Haskell (www.haskell.org)? Free, and it has
unlimited integers and all operation needed.

What is it about Haskell that allows it to transcend the storage
limitations of the entire known universe?

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Richard Heathfield <http://www.cpax.org.uk>
Email: -http://www. +rjh@
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