OpenEuphoria

Re: De-allocating memory - time issue

• Posted by Robert Craig <rds at RapidEuphoria.com> Sep 21, 2002
• 819 views

rforno writes:
> t = repeat(repeat(repeat(1.13218966051, 3000), 100), 50)

After executing this (type of) statement, in almost 
any language but Euphoria, you would have 3000x100x50 = 15 million 
(8-byte or more) copies of 1.13218966051 stored in memory. 
In Euphoria, you will have 3000 4-byte pointers to a *single* 
copy of 1.13218966051. You will also have 100 4-byte pointers 
to a *single* copy of the 3000-element sequence. You will also 
have 50 4-byte pointers to a *single* copy of  the 100-element sequence.
Instead of using 15 million x 8 bytes = 120,000,000 bytes of storage,
Euphoria will use only 3000x4+100x4+50x4 = 12,600 bytes 
plus the storage for *one* copy of 1.13218966051.

> for i = 1 to 50 do
>    for j = 1 to 100 do
> for k = 1 to 3000 do
>     t[i][j][k] = 1.31764903217
> end for
>    end for
> end for

During execution of the above nested loops,
copies will be made of all the sequences at each level.
There will now be 15 million 4-byte pointers 
to a *single* copy of 1.31764903217.
So you are still using much less memory than other languages.

> But if I perform a floating point calculation like:
> t[i][j][k] = (i * j * k) * 1.31764903217
> it lasts a long while 

Now it's no longer possible to have lots of pointers
to a single floating-point value. You are creating a
different f.p. value to store in each element. So you 
have created 15 million new floating-point values.
Each floating-point value needs 8 bytes plus 4 bytes
for a reference count, plus 4 bytes for malloc overhead.
So you suddenly need 15 million x 16 = 240,000,000 bytes
of *extra* storage in this case. Unless you have a lot of RAM
on your machine, you will be faced with some very serious
paging activity that will destroy your performance.

Regards,
   Rob Craig
   Rapid Deployment Software
   http://www.RapidEuphoria.com