$retval = $number > 16 ? 16 : int($number)
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Update: but see int for the implications of this construct. Basically, you should probably use POSIX::floor instead of int().

updated see reply below

I think you mean
$retval = $number > 16 ? 16 : int($number)
since yours would return 16 every time?

int() might return 0, given a value less than 1, though, so POSIX::floor() might still be a better plan.

Unless I'm grossly misunderstanding your question, you should give your function a better name, as "round" implies a well-known set of behaviors: round up, round down, round nearest. With your extra requirements, a future maintainer could easily be led astray by the name.

From your description, you want a number to be rounded down to the nearest integer with the additional requirements of:

(1) never returning a number greater than 16, and
(2) never returning 0.

So RoundDownSpecial might be a better name.

Also, you didn't specify what should be returned instead of 0 in the cases that normal rounding down would return 0. Other requirements are also unspecified, so you could get an entire zoo of behaviors. My candidate for your zoo:

#!/usr/bin/perl -w
use strict;
use warnings;

sub RoundDownSpecial_bizarro {
        my $num = shift;
        my $den = shift;

        return 7 unless $den+0.0;
        return 8 unless $num+0.0;

        my $tmp = int(($num+0.0) / $den);
        return 16 if $tmp > 16;
        return 16 if $tmp < 1;

        return int($tmp);
}

while (<DATA>) {
        chomp;
        my ($n, $d) = split /,/;
        print join(" ",$n, $d, RoundDownSpecial_bizarro($n, $d)), "\n"
+;
}

__DATA__
0, 1
10, 0
100, 1
1, 100
2, 1
8, 1
15, 1
16, 1
17, 1
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Mechanical Pedant

UPDATE: Apparently I'm not quite pedantic enough this morning. I missed some parenthesis around the '$num+0.0' clause. Thanks for the catch, FunkyMonk!

$reasult = int($number1/$number2);
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