integer

       point

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

DESCRIPTION
       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 in performance.

       Note that this only affects how most of the arithmetic and relational
       operators handle their operands and results, and not how all numbers
       everywhere are treated.  Specifically, "use integer;" has the effect
       that before computing the results of the arithmetic operators (+, -, *,
       /, %, +=, -=, *=, /=, %=, and unary minus), the comparison operators
       (<, <=, >, >=, ==, !=, <=>), and the bitwise operators (|, &, ^, <<,
       >>, |=, &=, ^=, <<=, >>=), the operands have their fractional portions
       truncated (or floored), and the result will have its fractional portion
       truncated as well.  In addition, the range of operands and results is
       restricted to that of familiar two's complement integers, i.e.,
       -(2**31) .. (2**31-1) on 32-bit architectures, and -(2**63) ..
       (2**63-1) on 64-bit architectures.  For example, this code

           use integer;
           $x = 5.8;
           $y = 2.5;
           $z = 2.7;
           $a = 2**31 - 1;  # Largest positive integer on 32-bit machines
           $, = ", ";
           print $x, -$x, $x + $y, $x - $y, $x / $y, $x * $y, $y == $z, $a, $a + 1;

       will print:  5.8, -5, 7, 3, 2, 10, 1, 2147483647, -2147483648

       Note that $x is still printed as having its true non-integer value of
       5.8 since it wasn't operated on.  And note too the wrap-around from the
       largest positive integer to the largest negative one.   Also, arguments
       passed to functions and the values returned by them are not affected by
       "use integer;".  E.g.,

           srand(1.5);
           $, = ", ";
           print sin(.5), cos(.5), atan2(1,2), sqrt(2), rand(10);

       will give the same result with or without "use integer;"  The power
       operator "**" is also not affected, so that 2 ** .5 is always the
       square root of 2.  Now, it so happens that the pre- and post- increment
       and decrement operators, ++ and --, are not affected by "use integer;"
       either.  Some may rightly consider this to be a bug -- but at least
       it's a long-standing one.

       Finally, "use integer;" also has an additional affect on the bitwise
           % perl -Minteger -le 'print (4 % -3)'
           1

       See "Pragmatic Modules" in perlmodlib, "Integer Arithmetic" in perlop



perl v5.14.2                      2010-12-30                    integer(3perl)
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