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

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In reply to Re^3: Question on simple arithmetic using perl by Marshall
in thread Question on simple arithmetic using perl by perlbaski

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