lt versus le in string comparision

Len has asked for the wisdom of the Perl Monks concerning the following question: ⭐ (strings)

Can anyone explain me this:

$char = 'a';

while ($char lt 'z')
{
   print $char;
   $char++;
}
[download]

prints as expected the characters (a..y) but if I change lt in le in the while expression:

while ( $char le 'z' )
[download]

it prints (a..yz) instead of the expected (a..z)

Originally posted as a Categorized Question.

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Re: lt versus le in string comparision⭐ by Abigail-II (Bishop) on Jun 13, 2002 at 14:06 UTC
Consider this: `my $str = 'y'; $str ++; print "$str\n"; $str ++; print "$str\n";` [download] This will print z aa Now, both `z` and `aa` are less or equal to `z`. It isn't until you reach `za` that you have a string that isn't less or equal to `z`. Abigail	[reply] [d/l]