Re: spinoza programming language and big numbers

Three Headed Monkey said:

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.

He is wrong. It won't. It *may* have numbers upon which the language places
no limitation, but the universe will impose its own limit.

Can problem be expressed in spinoza programming language?

Expressing the problem is not the problem. The size of the answer is the

Where can I buy / download spinoza compiler?

You can't buy or download it, but you can run it whenever you like. You
don't need to supply a program, because the compiler works out the program
you ought to be writing, and then compiles it for you, turning your vague
ideas about data processing into true spinoza-style output. To get this
wonderful functionality, visit: <>

Richard Heathfield <>
Email: -http://www. +rjh@
Google users: <>
"Usenet is a strange place" - dmr 29 July 1999