idiom for ( 1 < $foo < 10) ?

[tenya]

Is there an idiom for this:

(1 < $foo < 10) or die;

Edit 2001-05-18 by mirod: changed the title from "idiom for" to "idiom for ( 1 < $foo < 10) ?"

[jynx]

A common way to do it? How about this:

(1 < $foo && $foo < 10) or die "$foo not in range:";

Did you have something else in mind?

HTH,
jynx

[MeowChow]

This topic was recently discussed in this thread.

Oh, you want an idiom? How about...

grep int $foo == $_, 1..9 or die;

Just kidding :-)

And yes, I know, this screws up the 1 < x part...

   MeowChow                                   
               s aamecha.s a..a\u$&owag.print

[damian1301]

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!");

UPDATE:Spaced it out for clarity...and because I am/was bored.

Tiptoeing up to a Perl hacker.
Dave AKA damian

[clemburg]

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" ;

Christian Lemburg
Brainbench MVP for Perl
http://www.brainbench.com