Author: lwall
Date: 2010-02-20 06:02:01 +0100 (Sat, 20 Feb 2010)
New Revision: 29787
Modified:
docs/Perl6/Spec/S03-operators.pod
Log:
[S03] upon further reflection, we can make eternaleye++'s conjecture work right.
Modified: docs/Perl6/Spec/S03-operators.pod
===================================================================
--- docs/Perl6/Spec/S03-operators.pod 2010-02-20 04:48:40 UTC (rev 29786)
+++ docs/Perl6/Spec/S03-operators.pod 2010-02-20 05:02:01 UTC (rev 29787)
@@ -1995,8 +1995,19 @@
100, 200, 300 ... 900
since an exactly matching limit is returned as part of the
-sequence.
+sequence. And, in fact, since C<...> is list associative,
+and the heuristic depends only on the list to the immediate
+left, we can even say:
+ 1, 2 ...
+ 10, 20 ...
+ 100, 200 ... 900
+
+This works because the second C<...> sees only the 10,20, not
+the 9 before that, and likewise the third C<...> is blind to
+the 90 value. You can use parens to force one C<...> to be
+part of the list of another C<...> operator.
+
The exact function deduced depends on the direction from the final
value on the left to the limit value on the right. If the limit is
greater than the last value according to C<cmp>, then comparisons
