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.
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