My first thought when I saw this question was sprintf, which is Perl's standard approach to rounding, but then I saw ghenry's comment and realized that sprintf doesn't do what the OP specifically asked for, because for numbers exactly at the halfway point between two integers, perl always rounds towards the even number. So, as ghenry notes, sprintf '%.f', 28.5 yields 28, not the 29 that the OP requested.

Therefore, I think that Ted Pride's solution is the one that best meets the OP's requirement, since it will always round up at the halfway points.

But keep in mind that with this solution, a negative halfway-point number such as -28.5 gets rounded to -28, not -29.

I'm not sure what the best solution would be if what one wants is that halfway-point numbers get rounded "away from zero" (as opposed to "up"), so that -28.5 gets rounded to -29. One awkward solution is to mess around with absolute values and signs, like this: 

sub round {
  my $n = shift;
  my $abs  = abs( $n );
  return 0 if $abs < 0.5;
  my $sign = $n/$abs;
  return int( $abs + 0.5 ) * ( $n/$abs );
}
print round( $_ ), $/ for qw( -1.5 -0.5 0 0.5 1.5 );
__END__
-2
-1
0
1
2

[download]
But, eeewwww! I hope there's a more elegant approach...

Update: Replaced the test for the 0 condition, to avoid divisions by very small numbers. But this modest improvement is moot, since Roy Johnson's solution++ is indeed far more palatable. I had not realized that one can get a nice sign operator out of <=>. That alone takes care of my Perl learning quota for today...

the lowliest monk

In reply to Re: Rounding a number? by tlm
in thread Rounding a number? by Anonymous Monk

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