Roman Makurin wrote:
> On Wed, Jun 24, 2009 at 03:25:57PM +0200, Jenda Krynicky wrote:
>> From: Roman Makurin <dro...@gmail.com>
>>> here is complite perl script which produces such results without
>>> any warning:
>>>
>>> #!/usr/bin/perl
>>>
>>> use strict;
>>> use warnings;
>>>
>>> use constant {
>>> A => 0,
>>> B => 1,
>>> C => 2 };
>>>
>>> my @a = (A, B, C);
>>> my @b = (1, 2, 3);
>>>
>>> while(my $i = shift @a) {
>>> print $i, $/
>>> }
>> But of course this does not print anything. The shift(@a) returns the
>> first element of @a which is zero, assigns that to $i and then checks
>> whether it's true. And of course it's not. So it skips the body and
>> leaves the loop. Keep in mind that the value of
>>
>> my $i = shift @a
>>
>> is NOT a true/false whether there was something shifted from the
>> array. It's the value that was removed from the array and assigned to
>> the $i. And if that value it false (undef, 0, 0.0, "0", "0.0", "" -
>> if I remember rigth) then the whole expression evaluates to false in
>> boolean context.If I understand correctly, what you are saying is that while() is evaluating the left side of the '=' as it's condition, culminating into: while($i) Which eventually equates into: while(0) ...on the very first pass. >> Whether you use constants or not is irrelevant. You'd see the same >> behaviour with >> >> my @a = (0, 1, 2); > > big thanks for explanation :) +1. Steve
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