idiom for ( 1 < $foo

tenya has asked for the wisdom of the Perl Monks concerning the following question:

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Re: idiom for by jynx (Priest) on May 18, 2001 at 06:20 UTC
A common way to do it? How about this: `(1 < $foo && $foo < 10) or die "$foo not in range:";` [download] Did you have something else in mind? HTH, jynx	[reply] [d/l]
Re: idiom for by MeowChow (Vicar) on May 18, 2001 at 07:35 UTC
This topic was recently discussed in this thread. Oh, you want an idiom? How about... `grep int $foo == $_, 1..9 or die;` [download] Just kidding :-) And yes, I know, this screws up the `1 < x` part... MeowChow s aamecha.s a..a\u$&owag.print	[reply] [d/l] [select]
Re: idiom for by damian1301 (Curate) on May 18, 2001 at 06:27 UTC
or if you want to do something if that results to true, you can do this: `(1 < $foo && $foo < 10) ? do_stuff($foo) : die("nooo!");` [download] UPDATE:Spaced it out for clarity...and because I am/was bored. Tiptoeing up to a Perl hacker. Dave AKA damian	[reply] [d/l]
Re: idiom for by clemburg (Curate) on May 18, 2001 at 12:56 UTC
Repost from earlier thread: sub is_ordered_according_to { my ($binary_predicate, @values) = @_; if (@values < 3) { return $binary_predicate->(@values); } else { return $binary_predicate->($values[0], $values[1]) && is_ordered_according_to($binary_predicate, @values[1..$#values]); } } sub less_than { return shift() < shift(); } print is_ordered_according_to(\&less_than, 1, 2, 3, 4, 5, 6) ? "yes" : "no", "\n" ; print is_ordered_according_to(\&less_than, 3, 2, 1, 4, 5, 6) ? "yes" : "no", "\n" ; print is_ordered_according_to(\&less_than, 1, 2, 3, 4, 6, 5) ? "yes" : "no", "\n" ; [download] Christian Lemburg Brainbench MVP for Perl http://www.brainbench.com	[reply] [d/l]