$foo = 1500;
printf ("%.2f", $foo/100);
[download]

#!/usr/bin/perl -w

use strict;

my $foo;

$foo = 1500;
print myformat($foo, 2) . "\n";

sub myformat {
    my ($number, $places) = @_;
    my $decimal;

    $decimal = "";
    for (1 .. $places) {
        if ($number) {
            $decimal .= chop $number;
        }
        else {
            $decimal .= 0;
        }
    }
    return "$number\." . reverse $decimal;
}
[download]

I also added the reverse because I no-so-brillantly forgot that the decimals would be reversed.

Rich

Your plucking the digits off one end, sticking on the dot, and sticking them back on sounds like something a regex should do!

$number .= '0'  while length($number) < $places;
  # add trailing zeros if needed
$number =~ /(.{$places})/\.$1/;
[download]

Or a substr

# add trailing zeros, as above.
substr ($number, -$places, 0, '.');
[download]

Acutally a regex isn't all that good of an idea. It's very slow. But substr is better than chop.. benchmarks included..

Benchmark: timing 100000 iterations of chop, math, regex, substr...
      chop:  6 wallclock secs ( 4.92 usr +  0.03 sys = 4.95 CPU) @ 202
+02.02/s (n=100000)
      math:  2 wallclock secs ( 1.07 usr +  0.02 sys = 1.09 CPU) @ 917
+43.12/s (n=100000)
     regex: 25 wallclock secs (17.40 usr +  0.32 sys = 17.72 CPU) @ 56
+43.34/s (n=100000)
    substr:  2 wallclock secs ( 1.99 usr +  0.03 sys = 2.02 CPU) @ 495
+04.95/s (n=100000)
[download]

Doing the math is still the best option..

First of all, let me say that there must be a better way for you to store your data. However, you might be stuck with the format, for all I know, so on with an attempted solution ...

Okay, let's say you've split one line into fields using the appropriate module. You store those fields in @currentLine. Then you can do something like:


foreach (@currentLine[3 .. 11]) {
   printf ("%.*f", $currentLine[2], $_/(10 ** $currentLine[2])
}
[download]

The * sucks up the next variable in the list and uses it as the field width; in this case, it slurps the first appearance of $currentLine[2] in the list.

Update: As tilly has noted, the module I mentioned above does not handle newlines embedded within records. Another module, recommended by the same esteemed personage who is on vacation does that, if it is required.

-HZ

You're right about the data. The file is given to me by a client. I have no control over the format.

Then you just want to do a..

#!/usr/bin/perl -w

use strict;

my ($foo, $bar);

$foo = 1500;
$bar = 2;

print $foo / (10 ** $bar) ."\n";
[download]

Hope this helps..
Rich

First make sure $bar is numeric, then...

printf ("%.${bar}f", $foo / (10 ** $bar));
[download]

To deconstruct:

Take 10 to the power $bar to get the correct divisor. Then divide by it to move the decimal place over non-destructively. Then, to give the correct number of places, use printf and %. f -- note that there must be {} around bar because otherwise Perl will try to interpolate $barf, which, if you're using strict and have no variable called $barf will indeed barf. To get the string value rather than printing it out, use sprintf rather than printf.

So, for instance, if you have $foo being 570 and $bar being 1, you get (reducing the expression, so to speak):

1. printf ("%.${bar}f", $foo / (10 ** $bar))

2. printf ("%.1f", 570 / (10 ** 1))

3. printf ("%.1f", 570 / 10)

4. printf ("%.1f", 57)

5. "57.0"

... which is what you wanted I think.

Do you want the decimals even if they aren't siginficant? 2 zeros aren't significant. if you change the number to 1555, it will print 15.55. It just drops non significant digits.

yes I do. I understand 15 is the same as 15.00, but I need these zeros to remain in $foo's end result. The original digits in $foo must remain (i.e. $foo=1500 then result must have 1500 in it).
I'm sorry, I did not try to use a different number in your script.

#!/usr/bin/perl -w
 
use strict;
 
my $foo = 17432;
my $bar = 3;
 
$foo =~ s/([0-9]+)([0-9]{$bar})/$1.$2/; 
print $foo;
[download]

MP