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size of Centaur
- Subject: size of Centaur
- From: td@capa.inria.fr (Thierry Despeyroux)
- Date: 5 Dec 91 09:41:12 GMT
Nicos asked me the following question. As the answer is of general
interest, I will give the answer in this mailing-list...
=> Finally another question in a previous message on the centaurlist
=>you said that the parameter given to CENTAUR corresponded to a 64K area of
=>lists, so centaur 10 corresponds to 640K. This also applies to
=>CENTAUR 1.1 ?. In the previous answer you said:
=>
=>> The size of your Centaur is ~22Megs. It is a hudge system, and uses
=>> most of your 24Megs memory.
=>
=>This doesn't match with the previous answer: 90 * 64 = 5760 K. Is maybe the
=>new conversion 256K?
Le_Lisp uses different areas to store different objects.
The size of the process is the sum of these areas plus the size
of the kernel. Only the size of the CONS area may be given
when calling Centaur. The other are fixed in the configuration file.
Here is the description of thesee areas, as given in the Le_Lisp manual:
name unit real size
STACK k-word 1 = 4k-byte
CODE k-byte 1 = 1k-byte
HEAP k-byte 1 = 1k-byte
VECTOR k-vector 1 = 1k-vector = 8k-byte
NUMBER 0 on current systems
FLOAT k-float 1 = 1k-float = 8 or 0 k-byte
(depends on the version of Lisp)
STRING k-string 1 = 1k-string = 8k-byte
SYMBOL k-symbol 1 = 1k-symbol = 64k-byte
CONS 8k-cons 1 = 8k-cons = 64k-byte
The Mu-Prolog stack takes place in the Lisp CODE area. This is why this
area must be adapted statically.
We are making experiments with a two-processes Centaur in which Lisp and
Prolog will run in two different processes. It will be possible to change
the sizes of the prolog system without reconfiguring Centaur itself.
--
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