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-rw-r--r--Porting/perl5160delta.pod10
1 files changed, 5 insertions, 5 deletions
diff --git a/Porting/perl5160delta.pod b/Porting/perl5160delta.pod
index ca3893c0ba..04586855fc 100644
--- a/Porting/perl5160delta.pod
+++ b/Porting/perl5160delta.pod
@@ -3261,11 +3261,11 @@ working for the rest of the block.t
For list assignments like C<($a,$b) = ($b,$a)>, Perl has to make a copy of
the items on the right-hand side before assignment them to the left. For
efficiency's sake, it assigns the values on the right straight to the items
-on the left no variable is mentioned on both sides, as in
-C<($a,$b) = ($c,$d)>. The logic for determining when it can cheat was
-faulty, in that C<&&> and C<||> on the right-hand side could fool it. So
-C<($a,$b) = $some_true_value && ($b,$a)> would end up assigning the value
-of C<$b> to both scalars.
+on the left if no one variable is mentioned on both sides, as in C<($a,$b) =
+($c,$d)>. The logic for determining when it can cheat was faulty, in that
+C<&&> and C<||> on the right-hand side could fool it. So C<($a,$b) =
+$some_true_value && ($b,$a)> would end up assigning the value of C<$b> to
+both scalars.
=item *