How should I do integer division in Perl?

There are at least 2 reasonable answers to this question. (I originally gave only answer 2.)

1. Use the int() function to truncate the floating-point calculation result to an integer (throwing away the decimal part), as Bryan suggested in his self-answer: #539805

2. Use the use integer pragma to make Perl truncate both the inputs and results of calculations to integers. It's scoped to within { } blocks.

Examples:

print 3.0/2.1 . "\n";      # => 1.42857142857143
print 5.0/1.5 . "\n";      # => 3.33333333333333

print int(3.0/2.1) . "\n"; # => 1
print int(5.0/1.5) . "\n"; # => 3

{
  use integer;
  print 3.0/2.1 . "\n";    # => 1
  print 5.0/1.5 . "\n";    # => 5 (because 1.5 was truncated to 1)
}
print 3.0/2.1 . "\n";      # => 1.42857142857143 again

You can cast ints in Perl:

int(5/1.5) = 3;

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.