in reply to Question on simple arithmetic using perl

When the number becomes too large for an integer it is automagically promoted to a float.

  • Comment on Re: Question on simple arithmetic using perl

Replies are listed 'Best First'.
Re^2: Question on simple arithmetic using perl
by perlbaski (Sexton) on Sep 28, 2011 at 21:43 UTC
    Thank you Monk!! So does this mean its fine to use such arithmetic and not worry about whether perl can handle it correctly?
[reply]
      No. A float can represent a wider range of values, but a lesser number of significant digits than an integer.

      I'm not quite sure what the question is: 

      #!/usr/bin/perl -w
use strict;

my $all_bits_are_on = -1;
printf ("all_bits => %X\n", $all_bits_are_one);

my $x = $all_bits_are_on + 1;
print "Adding -1 to +1 is: $x\n";

__END__
all_bits => FFFFFFFF
Adding -1 to +1 is: 0

      [download]
[reply]
[d/l]

        Note that perl is actually using long and double. On 32bit double has more significant digits than long.

[reply]

In Section Seekers of Perl Wisdom