# FAQ 4.4 Does Perl have a round() function? What about ceil() and floor()? Trig functions...

#### Do you have a question? Post it now! No Registration Necessary.  Now with pictures!

•  Subject
• Author
• Posted on
comes with the standard Perl distribution. These postings aim to
reduce the number of repeated questions as well as allow the community
perlfaq is at http://faq.perl.org .

--------------------------------------------------------------------

4.4: Does Perl have a round() function?  What about ceil() and floor()?  Trig
functions?

Remember that "int()" merely truncates toward 0. For rounding to a
certain number of digits, "sprintf()" or "printf()" is usually the
easiest route.

printf("%.3f", 3.1415926535);   # prints 3.142

The "POSIX" module (part of the standard Perl distribution) implements
"ceil()", "floor()", and a number of other mathematical and
trigonometric functions.

use POSIX;
\$ceil   = ceil(3.5);   # 4
\$floor  = floor(3.5);  # 3

In 5.000 to 5.003 perls, trigonometry was done in the "Math::Complex"
module. With 5.004, the "Math::Trig" module (part of the standard Perl
distribution) implements the trigonometric functions. Internally it uses
the "Math::Complex" module and some functions can break out from the
real axis into the complex plane, for example the inverse sine of 2.

Rounding in financial applications can have serious implications, and
the rounding method used should be specified precisely. In these cases,
it probably pays not to trust whichever system rounding is being used by
Perl, but to instead implement the rounding function you need yourself.

To see why, notice how you'll still have an issue on half-way-point
alternation:

for (\$i = 0; \$i < 1.01; \$i += 0.05) { printf "%.1f ",\$i}

0.0 0.1 0.1 0.2 0.2 0.2 0.3 0.3 0.4 0.4 0.5 0.5 0.6 0.7 0.7
0.8 0.8 0.9 0.9 1.0 1.0

Don't blame Perl. It's the same as in C. IEEE says we have to do this.
Perl numbers whose absolute values are integers under 2**31 (on 32 bit
machines) will work pretty much like mathematical integers. Other
numbers are not guaranteed.

--------------------------------------------------------------------

The perlfaq-workers, a group of volunteers, maintain the perlfaq. They
are not necessarily experts in every domain where Perl might show up,
so please include as much information as possible and relevant in any
operating system or platform, so please include relevant details for
corrections to examples that do not work on particular platforms.
Working code is greatly appreciated.

If you'd like to help maintain the perlfaq, see the details in
perlfaq.pod.

## Re: FAQ 4.4 Does Perl have a round() function? What about ceil() and floor()? Trig functions?

That example conflates two different issues:

1) 0.05 is not exactly representable as a binary floating point number,
so \$i doesn't get the values 0.00, 0.05, 0.10, 0.15, 0.20, 0.25, ...
as one might naively expect. Instead it gets the values
0.00000000000000000000
0.05000000000000000278
0.10000000000000000555
0.15000000000000002220
0.20000000000000001110
0.25000000000000000000
...
(rounded to 20 digits after the comma). So the second value
(0.05000000000000000278) is actually a bit above 0.05 and therefore
has to be rounded up according to the "round to nearest" rule.
So most of these numbers are not at the "half-way-point", and can be
clearly and anambiguosly rounded (but the result is not what you
expect when you think they are at the half-way-point.

2) The one exception is 0.25 which just happens to be exact because all
the rounding errors cancel out at that point[1]. 0.25 is exactly
half-way between 0.2 and 0.3 so "round to nearest" cannot decide
between these two. In school we all learned to round up in this case.
The IEEE rules, however, mandate that one must round to the nearest
*even* number, i.e., 0.2 in this case.

I'm not sure which of these issues the example should demonstrate, but
showing both in a single example without adequate explanation is
confusing. The "half-way-point" problem can be demonstrated correctly
with:

for (\$i = 0; \$i <= 10; \$i += 0.5) { printf "%.0f ",\$i}

0 0 1 2 2 2 3 4 4 4 5 6 6 6 7 8 8 8 9 10 10

The fact that decimal fractions are not generally representable in
binary fp and that adding them is an especially bad idea can be
demonstrated better with:

for (\$i = 0, \$j = 0; \$i < 1.01; \$i += 0.05, \$j += 5) {
print \$i * 100 - \$j, " ";
}

0 0 0 1.77635683940025e-15 0 0 0 0 0 -7.105427357601e-15
-7.105427357601e-15 -7.105427357601e-15 0 0 0
1.4210854715202e-14 1.4210854715202e-14 1.4210854715202e-14
2.8421709430404e-14 2.8421709430404e-14 2.8421709430404e-14

(Although there are still too many 0s in the output - maybe somebody can
find a better example).

2**53 actually, if the machine uses IEEE FP arithmetic.

hp

[1] I didn't expect that and I even repeated the calculation with pencil
and paper. Adding a number which is slightly larger than 0.05 5 times
gives exactly 0.25 in IEEE double arithmetic.