convert number to format


I need to convert numbers like .7 and 2.3 to 00001 and 00011 respectivly.

0.7 is to be rounded to 1 which is to be 00001
2.3 is to be rounded to 2 which is to be 00011
4.5 is to be rounded to 5 which is to be 11111

so here is my best shut
my $results = flt2bit5 ( .7);

sub flt2bit5 {
    my $n = shift;
    my $n = sprintf ("%.0f", $n);
I am not sure ....

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This does the trick:

sub flt2bit5 {
   my $n = shift;
   my $res;
   for( my $i=5-int($n+0.5); $i>0; $i-- ) {
      $res .= "0";
   for( $i=0; $i<int($n+0.5); $i++ ) {
      $res .= "1";
   return $res;

int() is basically a floor() function, so by adding 0.5 you get an actual round(). You have to fill with 0 for 5-n digits and then print n times 1.

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You could also use the string mulitplication operator..

sub flt2bit5 {
   my ($n) = @_;
   $n = int ($n + 0.5);
   $n = 5 if ($n > 5);  # Just in case
   $n = 0 if ($n < 0);  # Just in case
   return "0" x (5 - $n) . "1" x $n;
2.3 is to be rounded to 2 which is to be 00011

That should be 00010

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If you're tring to convert to binary, then this will work.

use strict;

my @num = qw(0.7 2.3 4.5);

foreach (@num) {
   my $results = flt2bit5($_);
   print "$_ is to be rounded to ".sprintf ("%.0f", $_)." which is to be $results\n";

sub flt2bit5 {
   my $n = sprintf ("%.0f", shift);
   return sprintf ("%.5b", $n);
If you want the output to be as you posted, then change the return statement to this:

return sprintf ("%0.5d", 1 x $n);
samjAuthor Commented:
thanks for to all for all the comments, arjanh had the fastest code when I benchmarked them.

thanks again
FishMonger has it.  You won't even need the function call, just change it to:

my $results = spirntf("%0.5d", "1" x ($input + 0.5));

If inputut is less than 0, you will still get "00000", but if $input is > 5, you will get a field more than 5 characters.  Even the error checking can be done in one line...

my $results = sprintf("%0.5d", "1" x (($input > 4.5) ? 5 : $input + 0.5));
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