Re: how to always round up in perl?

[philipbailey]

in reply to how to always round up in perl?

Or if you do not want to import a mass of functions into your namespace from the POSIX module, just do this:

my $n = 3.3;
my $ceiling = int($n + 1);

(Update) Above is incorrect, as pointed out by others. This works:

my $ceiling = ($n == int $n) ? $n : int($n + 1);

[Ratazong]

Which I wouldn't recommend, because it fails in the following situation (with $n2) ...

my $n1 = 3.3;
my $n2 = 4;
print int($n1 + 1)," ",int($n2 + 1), "\n";

[chrestomanci]

That fails if $n is allready an integer. You would need something like:

my $n = 3.0;
my $ceiling = int($n);
$ceiling += 1 if $ceiling != $n;

Or as a one liner:

my $ceiling = int($n) + ($n != int($n));

[toolic]

if you do not want to import a mass of functions into your namespace from the POSIX module

... just do this:

use warnings; 
use strict; 
use POSIX qw(ceil);

for (3.3, 3.6, 3003.3) {
    print $_, "\t", ceil($_), "\n";
}

__END__

3.3    4
3.6    4
3003.3    3004

According to How to Import, this selectively imports only the ceil function.

Alternately, you could import nothing and call the function using POSIX::ceil.