Re: OT: Integer rounding?

You might get some help from this Q&A.

Update:
Doing it by hand, I'd do:

sub round_to_nearest {
    my ($n, $scale) = @_;
    int(($n + $scale/2)/$scale) * $scale;
}
[download]

The PerlMonk tr/// Advocate

Comment on Re: OT: Integer rounding? Download Code

Replies are listed 'Best First'.
Re: Re: OT: Integer rounding? by flyingmoose (Priest) on Mar 22, 2004 at 21:43 UTC
A few sources: perlfaq4 `sub round { my($number) = shift; return int($number + .5); }` [download] perldoc -f int int Returns the integer portion of EXPR. If EXPR is omitted, uses "$_". You should not use this function for rounding: one because it truncates towards "0", and two because machine representations of floating point numbers can sometimes produce counterintuitive results. For example, "int(-6.725/0.025)" produces -268 rather than the correct -269; that's because it's really more like -268.99999999999994315658 instead. Usually, the "sprintf", "printf", or the "POSIX::floor" and "POSIX::ceil" functions will serve you better than will int().	[reply] [d/l]

Replies are listed 'Best First'.

Re: Re: OT: Integer rounding?
by flyingmoose (Priest) on Mar 22, 2004 at 21:43 UTC

perlfaq4

sub round {
    my($number) = shift;
    return int($number + .5);
}
[download]

perldoc -f int

int Returns the integer portion of EXPR. If EXPR is omitted, uses "$_". You should not use this function for rounding: one because it truncates towards "0", and two because machine representations of floating point numbers can sometimes produce counterintuitive results. For example, "int(-6.725/0.025)" produces -268 rather than the correct -269; that's because it's really more like -268.99999999999994315658 instead. Usually, the "sprintf", "printf", or the "POSIX::floor" and "POSIX::ceil" functions will serve you better than will int().

[reply]
[d/l]