in reply to Why does it return 0?

sub dec2bin {
   my $inNumber = shift;
   my $sign = $inNumber=~s/^-// ? '-' : '';
   return $sign.sprintf "%b", $inNumber;
}

[download]
is how I'd do it (untested). (The "%b" format was added in perl 5.6.0.) Don't forget to actually use the parameter you pass. Also, in your code $binnumb = $binnumb . $binary would be usually written $binnumb .= $binary. (Either is exempt from the usual warning that $binnumb is undefined; other operators special-cased this way are +, -, ||, &&, ^, |, ++, --.)

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Re: Re: Why does it return 0?
by mAineAc (Novice) on Feb 05, 2004 at 02:17 UTC
    I tried to use the $binnumb .= $binary but it doesn't seem to be concatenating the string it just returns the last value of $binary.
[reply]
      In your original code, you needed to have had a loop, doing the % and / and append until the number is zero. But you're better off just using sprintf "%b" or unpack.

      Update: except that you'd have to do $binnumb = $binary . $binnumb instead, which will get a warning unless you initialize $binnumb to '' at the start. And you need to handle negative numbers.

[reply]
[d/l]
        Yea I realize I have to handle negative numbers still. I did get it to work with a while loop like this: 
        sub dec2bin {
    my $binnumb;
    while($inNumber > 0) {
    my $binary = $inNumber % 2;
    print $inNumber; print "\n";
    $inNumber = int($inNumber / 2);
    $binnumb .= $binary; 
    print $binnumb; print "\n";
    }
    $binnumb = reverse $binnumb;
    return $binnumb;
    }

        [download]
        I wasn't thinking when I tried to do it with an if statement. I don't get any errors with this and it gives me the right number. I just didn't understand the code that you gave me at first. What does %b do?
[reply]
[d/l]

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