Re: Representing decimal part as regular integer on low 16 bits?

Op Tue, 12 Sep 2006 23:32:15 +0200 schreef Frank Buss:

<snip>
A quick and dirty implementation in Forth:

A little bit too quick and too dirty I might add ;-)

\ reads a number with 3 digits
: n3 ( n n -- n) drop ;
With a 16 bit implementation you lose too much precision here!

\ multiplies two 3 digit numbers
: *3 ( n n -- n ) * 1000 / ;

Remember the */ operator that every Forth has, much better precision:
: *3 ( n n -- n ) 1000 */ ;


\ divides two 3 digit number
: /3 ( n n -- n ) swap 1000 * swap / ;

Here too:
: /3 ( n n -- n ) 1000 swap */ ;

1.234 n3 42.123 n3 *3 .3 -> 51.979
Does not do what you want in a 16 bit Forth.

51.979 n3 42.123 n3 /3 .3 -> 1.233

Is it possible to write a parser for "n3" in Forth, which reads the next
number sequence like "create" reads the next char sequence? I've read
something about >in, source etc., but don't have an idea how to use it.

In short, use doubles for 16 bit implementations: high word the integer
part, the low word as the fraction times 65536. Adding and subtracting with
the normal D+ and D-.

For 32 bit implementations you could use single precision and normal + and
-. Decimal point between bit 15 and 16.

--
Coos

CHForth, 16 bit DOS applications
http://home.hccnet.nl/j.j.haak/forth.html
.

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