Re: Rounding off numbers

[salva]

use POSIX 'floor';

sub round {
  my $x = shift;
  floor($x + 0.5);
}

You can use int instead of POSIX::floor if you don't need to round negative numbers. Anyway, read the docs for the int operator in perlfunc man page.

[Tanalis]

.. though it's important to remember that simply casting a float to an int, thus:

my $float = 2.744;
my $int   = int $float;

results in the decimal part of the number simply being truncated, giving an integer value with the above example of 2, not 3, which is possibly not what's expected.

Update: Although, of course, adding the 0.5 to the number solves the problem nicely. Didn't see that in the post until just now ..

[sgifford]

Adding 0.5 before the int, as in the above example, solves that.