Re: fun(?) with +=

Please note that:

$foo = shift || 1;
# is the same as
$foo = (shift || 1);
[download]

and

$foo = shift or 1;
# is the same as
($foo = shift) or 1;
[download]

see perldoc perlop about Operator Precedence

Comment on Re: fun(?) with += Select or Download Code

Replies are listed 'Best First'.
Re^2: fun(?) with += by Aristotle (Chancellor) on Jan 02, 2005 at 22:41 UTC
`$ perl -MO=Deparse,-p -e'$foo = shift \|\| 1' ($foo = (shift(@ARGV) \|\| 1)); -e syntax OK` [download] Makeshifts last the longest.	[reply] [d/l]