For details read Perl NOC. After June 25

Léon Brocard >
perl5.005_04 >
integer

integer - Perl pragma to compute arithmetic in integer instead of double

use integer; $x = 10/3; # $x is now 3, not 3.33333333333333333

This 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.

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