


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


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);




perl v5.10.1                2002-10-27                          1





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


       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 v5.10.1                2002-10-27                          2


