Re: [ID 20010305.005] "use integer" doesn't make rand() return integers

Message-ID: <Pine.SOL.3.91.1010306103113.9094A-100000@gateway.grumman.com>

use integer document clarification.

p4raw-id: //depot/perl@9115

diff --git a/lib/integer.pm b/lib/integer.pm
index f019fb3..f720ce5 100644 (file)
--- a/lib/integer.pm
+++ b/lib/integer.pm
@@ -4,7 +4,7 @@ our $VERSION = '1.00';
 
 =head1 NAME
 
-integer - Perl pragma to compute arithmetic in integer instead of double
+integer - Perl pragma to use integer arithmetic instead of floating point
 
 =head1 SYNOPSIS
 
@@ -14,32 +14,55 @@ integer - Perl pragma to compute arithmetic in integer instead of double
 
 =head1 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.
+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 affects the operations, not the numbers.  If you run this
-code
+Note that this only affects how certain operators handle their operands
+and results, and not all numbers everywhere.  Specifically, C<use
+integer;> has the effect that before computing the result of X + Y, X -
+Y, X / Y, X * Y, X % Y, or -X (unary minus), the operands X and Y have
+their fractional portions truncated, and the result will have its
+fractional portion truncated as well.  For example, this code
 
     use integer;
-    $x = 1.5;
-    $y = $x + 1;
-    $z = -1.5;
-
-you'll be left with C<$x == 1.5>, C<$y == 2> and C<$z == -1>.  The $z
-case happens because unary C<-> counts as an operation.
-
-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
+    $x = 5.8;
+    $y = 2.5;
+    $, = ", ";
+    print $x, -$x, $x + $y, $x - $y, $x / $y, $x * $y;
+
+will print:  5.8, -5, 7, 3, 2, 10
+
+Note that $x is still printed as having its true non-integer value of
+5.8 since it wasn't operated on.  Also, arguments passed to functions
+and the values returned by them are not affected by C<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 C<use integer;>  The power
+operator C<**> is also not affected, so that 2 ** .5 is always the
+square root of 2.
+
+Finally, C<use integer;> also has an affect on the bitwise operators
+"&", "|", "^", "~", "<<", and ">>".  Normally, the operands and results
+are treated as unsigned integers, but with C<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 L<perlmod/Pragmatic Modules>.