[perl] How should I do integer division in Perl?

What is a good way to always do integer division in Perl?

For example, I want:

real / int = int

int / real = int

int / int = int

int(x+.5) will round positive values toward the nearest integer. Rounding up is harder.

To round toward zero:

int($x)

For the solutions below, include the following statement:

use POSIX;

To round down: POSIX::floor($x)

To round up: POSIX::ceil($x)

To round away from zero: POSIX::floor($x) - int($x) + POSIX::ceil($x)

To round off to the nearest integer: POSIX::floor($x+.5)

Note that int($x+.5) fails badly for negative values. int(-2.1+.5) is int(-1.6), which is -1.

Eg 9 / 4 = 2.25

int(9) / int(4) = 2

9 / 4 - remainder / deniminator = 2

9 /4 - 9 % 4 / 4 = 2

you can:

use integer;

it is explained by Michael Ratanapintha or else use manually:

$a=3.7;
$b=2.1;

$c=int(int($a)/int($b));

notice, 'int' is not casting. this is function for converting number to integer form. this is because Perl 5 does not have separate integer division. exception is when you 'use integer'. Then you will lose real division.

Hope it works

int(9/4) = 2.

Thanks Manojkumar

The lexically scoped integer pragma forces Perl to use integer arithmetic in its scope:

print 3.0/2.1 . "\n";    # => 1.42857142857143
{
  use integer;
  print 3.0/2.1 . "\n";  # => 1
}
print 3.0/2.1 . "\n";    # => 1.42857142857143

Integer division $x divided by $y ...

$z = -1 & $x / $y

How does it work?

$x / $y

return the floating point division

&

perform a bit-wise AND

-1

stands for

&HFFFFFFFF

for the largest integer ... whence

$z = -1 & $x / $y

gives the integer division ...