spiccioni has asked for the wisdom of the Perl Monks concerning the following question:

Dear monks,
As a basic perl homework, I wrote the following script: 
$n=0;
print ($n,"\n");
while ($n<1){
    $n=$n+.1;
    print ($n,"\n");
}

[download]
The output is: 
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
1.1

[download]
Since the while condition is that n should be less than 1, I don't understand why the loop is entered for a last time when the value of n is 1, so that the value 1.1. is printed.
Thanks for any illumination.
Sara

Replies are listed 'Best First'.
Re: non-integer increments
by fglock (Vicar) on Dec 02, 2004 at 15:50 UTC

    Because floating point is not exact (again)

    $n=0;
print ($n,"\n");
while ($n<1){
    $n=$n+.1;
    printf ("%0.18f \n", $n);
}

    [download]
    0
0.100000000000000006 
0.200000000000000011 
0.300000000000000044 
0.400000000000000022 
0.500000000000000000 
0.599999999999999978 
0.699999999999999956 
0.799999999999999933 
0.899999999999999911 
0.999999999999999889 
1.099999999999999867

    [download]

    Each time you make a floating point addition, you are accumulating error.

[reply]
[d/l]
[select]
Re: non-integer increments
by xorl (Deacon) on Dec 02, 2004 at 15:54 UTC
    Once again I learn something from someone else's question. Now I have a question of my own. Is there some way to easily deal with this problem?
[reply]

      Yes, use an integer for the counter, and derive the floating point number from it. Integers are represented accurately and they don't accumulate error (unless they grow beyond a certain size).

      my $i=0;
print($i, "\n");
while ($i < 10){
    my $n = ++$i / 10;
    print($n, "\n");
}

      [download]

      Update: oops, typo fixed.

[reply]
[d/l]
        for (0..10) {
  print $n/10,"\n";
}

        [download]
        ---
        demerphq

[reply]
[d/l]
        just a small remark
        your probably meant
            $n = $i / 10;
        else you print out a few 0s
        si_lence
[reply]
[d/l]

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