integer(3pm)                                    Perl Programmers Reference Guide                                    integer(3pm)



NAME
       integer - Perl pragma to use integer arithmetic instead of floating 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 operators.  Normally, the operands and results are
       treated as unsigned integers, but with "use integer;" the operands and results are signed.  This means, among other
       things, that ~0 is -1, and -2 & -5 is -6.

       Internally, native integer arithmetic (as provided by your C compiler) is used.  This means that Perl's own semantics for
       arithmetic operations may not be preserved.  One common source of trouble is the modulus of negative numbers, which Perl
       does one way, but your hardware may do another.

           % perl -le 'print (4 % -3)'
           -2
           % perl -Minteger -le 'print (4 % -3)'
           1

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



perl v5.12.4                                               2011-06-01                                               integer(3pm)