Okay, very good. I stand corrected. (Well, mostly. I would pedantically argue that you may have arbitrary-precision arithmetic, but you don't have arbitrary-

Okay, very good. I stand corrected. (Well, mostly. I would pedantically argue that you may have arbitrary-precision arithmetic, but you don't have arbitrary-precision integers until they're a functional datatype, so you can write simply 'a + b' to add them -- with the automatic allocation that that in general requires. But this is a quibble.)