integer - Perl pragma to compute arithmetic in integer instead of double
use integer;
$x = 10/3;
# $x is now 3, not 3.33333333333333333This tells the compiler to use integer operations from here to the end of the enclosing BLOCK. On many machines, this doesn't matter a great deal for most computations, but on those without floating point hardware, it can make a big difference.
Note that this affects the operations, not the numbers. If you run this code
use integer;
$x = 1.5;
$y = $x + 1;
$z = -1.5;you'll be left with $x == 1.5, $y == 2 and $z == -1. The $z case happens because unary - counts as an operation.