pugs-comm...@feather.perl6.nl wrote:
Author: lwall
Date: 2010-01-23 15:43:40 +0100 (Sat, 23 Jan 2010)
New Revision: 29582
@@ -1191,8 +1191,6 @@
Block Callable
- Iterator List
- Seq Iterable
Range Iterable
Set Associative[Bool]
Bag Associative[UInt]
@@ -1225,17 +1223,14 @@
=head2 Mutable types
+ Iterator Perl list
+ Seq Partially or completely reified list
Scalar Perl scalar
Array Perl array
Hash Perl hash
@@ -1298,6 +1293,8 @@
+C<Seq> is mutable insofar as it it generated
+lazily, but the portion of the list that is already reified is
+considered immutable.
I'm not sure that declaring "Seq" mutable is the best move here.
I believe that the definitive test for considering something immutable or
mutable is in its semantics as observed by its consumers/readers.
If a Seq guarantees that multiple reads from it by anyone over time will see
exactly the same result, then it is immutable. Or put in functional terms, for
all Seq $s, if the result of "f($s)" is deterministic for every possible "f()",
then $s is immutable.
I believe that the fact Seq is lazy is just an implementation detail and has no
bearing on its immutable status as defined above. Sure, we may not know in
advance what all the elements of Seq will be until it is fully reified, but once
anyone tries to use those values once, any subsequent attempts by anyone to use
those values will get the *same* values, which is the point of immutability.
So I strongly recommend to make "Seq" immutable again.
Anything can be made lazy if the implementer wants to, and also anything can be
made mutable if Perl 6 permits access to the guts of that thing, such as to ask
"what are this Seq's reified elements" or "swap out the Seq's generator code
with this one", but allowing such suggests that all bets are off anyway.
-- Darren Duncan
