How do you round a floating point number in Perl?

Question

How can I round a decimal number (floating point) to the nearest integer?

e.g.

1.2 = 1
1.7 = 2

Answer 1

google: perl round -> perldoc/perlfaq

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.

Answer 2

cat table | perl -ne '/\d+\s+(\d+)\s+(\S+)/ && print "".int(log($1)/log(2))."\t$2\n";'

Answer 3

Whilst not disagreeing with the complex answers about half-way marks and so on, for the more common (and possibly trivial) use-case:

my $rounded = int($float + 0.5);