Casteele/ShadowLord wrote:
>
> Actually, Bill's answer has an error.. Adding .05 just introduces a subtle
> error because some
> decimal floating point numbers cannot be exactly represented in binary form.
> For example:
>
> my ($i, $j);
> foreach $i (0.5, 1.5, 2.5, 3.5, 4.5, 5.5, 6.5, 7.5, 8.5, 9.5) {
> $j = $i + .05;
> printf "%.20f\t%.20f\t%.1f\n", $i, $j, $j;
> }
That can be resolved by using a smaller value than .05. Actually, I use
something like this to do my rounding when CPU isn't important :
sub round ($;$) { # $ret = round ($num, [$digits:2])
my ($num, $dp) = (@_, 2); # default decimal places is 2
int ($num * 10 ** $dp + ($num < 0 ? -.5 : .5)) / 10 ** $dp;
}
so it would be :
foreach my $ii (0.5, 1.5, 2.5, 3.5, 4.5, 5.5, 6.5, 7.5, 8.5, 9.5) {
printf "%.0f\n", round ($ii);
}
Yielding:
0.5
1.5
2.5
3.5
4.5
5.5
6.5
7.5
8.5
9.5
> I think what Bill really meant to do.. To round a whole number (an integer),
> use something
> like this:
>
> my ($i, $j);
> foreach $i (0.5, 1.5, 2.5, 3.5, 4.5, 5.5, 6.5, 7.5, 8.5, 9.5) {
> $j = int($i + .5);
> printf "%.20f\t%.20f\t%.0f\n", $i, $j, $j;
> }
No, that's not what I intended to do. You can still use (s)printf alone
to do the job - just use something smaller than 1/2 of the next place (like
.01) or use a routine like the one above for something more robust. I
never use int in the process - but that's just me.
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