Author: larry
Date: Tue Feb 5 09:15:04 2008
New Revision: 14500
Modified:
doc/trunk/design/syn/S03.pod
Log:
Clarify that lhs of list assignment is list of containers, not thunks
Modified: doc/trunk/design/syn/S03.pod
==============================================================================
--- doc/trunk/design/syn/S03.pod (original)
+++ doc/trunk/design/syn/S03.pod Tue Feb 5 09:15:04 2008
@@ -12,9 +12,9 @@
Maintainer: Larry Wall <[EMAIL PROTECTED]>
Date: 8 Mar 2004
- Last Modified: 4 Feb 2008
+ Last Modified: 5 Feb 2008
Number: 3
- Version: 129
+ Version: 130
=head1 Overview
@@ -2021,7 +2021,7 @@
=item *
-The list (array) assignment operator now parses on the right like
+The list assignment operator now parses on the right like
any other list operator, so you don't need parens on the right side of:
@foo = 1, 2, 3;
@@ -2038,6 +2038,9 @@
($a, *, $c) = 1, 2, 3; # throw away the 2
($a, $b, $c, *) = 1..42; # throw away 4..42
+(Within signature syntax, a bare C<$> can ignore a single argument as well,
+and a bare C<*@> can ignore the remaining arguments.)
+
List assignment offers the list on the right to each container on the
left in turn, and each container may take one or more elements from the
front of the list. If there are any elements left over, a warning is
@@ -2058,9 +2061,17 @@
however, leave any subsequent lvalue containers with no elements,
just as in Perl 5.)
+The left side is evaluated completely for its sequence of containers before
+any assignment is done. Therefore this:
+
+ my $a = 0; my @b;
+ ($a, @b[$a]) = 1, 2;
+
+assigns 2 to @b[0], not @b[1].
+
=item *
-The item (scalar) assignment operator expects a single expression with
+The item assignment operator expects a single expression with
precedence tighter than comma, so
loop ($a = 1, $b = 2; ; $a++, $b++) {...}
@@ -2068,7 +2079,7 @@
works as a C programmer would expect. The term on the right of the
C<=> is always evaluated in item context.
-The syntactic distinction between scalar and list assignment is similar
+The syntactic distinction between item and list assignment is similar
to the way Perl 5 defines it, but has to be a little different because
we can no longer decide the nature of an inner subscript on the basis
of the outer sigil. So instead, item assignment is restricted to
