OpenEuphoria

Re: Referencing Nested Sequences

• Posted by Derek Parnell <ddparnell at bigpond.com> Jun 08, 2003
• 482 views

----- Original Message -----
From: <pblonner at optushome.com.au>
To: "EUforum" <EUforum at topica.com>
Subject: Re: Referencing Nested Sequences


>
>
> Thanks ,
>
> I did not expect such a quick response.
> Derek, I guess the ".au" postfix gave me away, I'm from Sydney.  Some of
the
> lingo you used might give us all away as a "wierd mob" :)

Well aren't we? 

> I realise that I may have simplified my example to where it lost some of
the
> general issue.   I was hoping/believing that Euphoria would be great for
> symbolic manipulations,i.e where you could coalesce sub-sequences into an
item
> that is part of a larger item,  generally rearrange items in a sequence -
where
> each of these items might be a nested sequence.
>
> Would this result in is that large 'copying tasks' for  each rearrangement
? -

Not really. If I understand this ( Robert Craig - can you confirm this?)
correctly, altering an element inside a sequence does not mean that the
entire sequence is copied to a new RAM location. Internally, a sequence uses
pointers to data, like a handle. This, combined with reference counting and
garbage collection, means that manipulating a sequence's contents is not all
that inefficient.


> even though it might just be pushing a item up or down the structure.
> That is, you may have just moved an subsequence - but as soon as you set a
value
> in the "moved" subseqence would Euphoria assume a copy of the entire
subsequence
> is necessary?


> This is perhaps why a sort library function is vital, since
> shifting the items manually around and then setting values later would
cause
> large 'copying' - i.e if the sort was coded outside the library function.

You can see the source code for the supplied sort() function in sort.e file.
It is really very quick.

Here is a short program to demonstrate that the length of a sequence does
not greatly influence the speed of changing its elements.

----
include machine.e
atom e
integer c,l
sequence x,y
tick_rate(1000)
x = repeat({0,1,2,3,4,5,6,7,8,9}, 4)
l=20
x = repeat(x,l)
y = {{"testdata"}}
c = 0
e = time()+10
while e > time() do
    x[2] = y
    c += 1
end while
? c
---

This just counts the number of assignments done in a 10 second span. Try
varying the value of 'l' and you should see that it doesn't change things
much.

BTW, on my machine I'm getting a count around 21 million.
--
Derek