OpenEuphoria

Re: New & question

• Posted by Robert Craig <rds at EMAIL.MSN.COM> Nov 10, 1998
• 619 views

Rolf Schroeder writes:
> I found once a subroutine from Rob (bytes_needed) from
> wich I concluded that the sequence s defined by
>        sequence s
>        s = repeat( 1.1, 1024)
> occupies 16 * 1024 bytes = 16 Kb (+ small overhead).
> Is that really true? An atom has an accuracy of 8 bytes,
> I would expect only 8Kb (+ small overhead) therefor.

s will occupy roughly 4 * 1024 = 4 Kb plus the space for *one*
floating-point atom. Why 4 *? Because s will consist of
1024 4-byte pointers (32-bit addresses) to a single
shared floating-point number (1.1) in memory. repeat()
will always set up shared pointers to the f.p. atom or
sequence that is repeated.

In general, floating-point atoms take more than 8 bytes, because
they also include a 4-byte reference count field, plus 4-bytes
for storage allocation overhead.

bytes_needed() assumes that no sharing takes place,
which is often wrong.

Regards,
     Rob Craig
     Rapid Deployment Software
     http://members.aol.com/FilesEu/