You might try the following code for padding numbers.

my $fill = "y";
my $len  = 8;
my $val  = 4040;

my $str  = substr (($fill x $len).$val, -$len, $len);

print ("$str\n");
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HTH, Rata

...or a loop :

my $str = 'barp';
$str = ">".$str until (length$str >=10); ## in case it is already more
+ than 10...
print $str . "\n";
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Obviously the loop could check for anything, the length of the pad, the number of a particular character, the time...

Updated code and added last bit

Just a something something...

I was wondering which that inventive method is

Switching to printf is far more than 4 characters.

print("$line_num $rest\n");
printf("%03d %s\n", $line_num, $rest);
                           12345678901
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Maybe he's referring to string autoincrement.

If you start with

$ perl -e'my $i=0; while (<>) { print ++$i, " $_" }' file
1 foo
2 bar
3 baz
4 qux
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then this is just four more characters:

                1 234
$ perl -e'my $i="000"; while (<>) { print ++$i, " $_" }' file
001 foo
002 bar
003 baz
004 qux
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(Note that the trick will fail if you use $i as an number.)

I would have started with

$ perl -e'while (<>) { print "$. $_" }' file
1 foo
2 bar
3 baz
4 qux
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so I would have had to add far more characters than four. In fact, using printf is still shorter:

$ perl -e'while (<>) { printf "%03d %s", $.,  $_ }' file
001 foo
002 bar
003 baz
004 qux
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(Avoiding -p and -l since this is suppose to be part of a larger program.)

print ++$i, " ...";
printf "%03d ...", ++$i;
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#!/usr/bin/perl -w
use strict;

printf("%03d\n", 5);
printf("%03d\n", 12345);
__END__
Prints:
005
12345
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I looked at your site reference http://www.comp.leeds.ac.uk/Perl/control.html#exercise and there is some really bogus advice there! Use chomp() to get rid of trailing LF or CR,LF, not chop()! chop() deletes the last char in the string no matter what it is! chomp() is conditional and only gets rid of trailing "new line" characters. In the case of Windows, chomp() will get rid of 2 characters (the CR and LF), on Unix only one (the LF) - in case of a simple string without any "new line" characters, nothing happens.

Anyway, please explain your application with something other than zero or space leading pad characters. That would be unusual and I'd like to understand it.

In the case of Windows, chomp() will get rid of 2 characters (the CR and LF)

Ah, yet another person fooled by the misleading statements in perlport, most likely.

chomp does not remove "\r" characters, even on Win32.

Actually, I would never use chomp. s/\s+$// is a much better idea. Not only does it remove "\r" characters when on Win32 but it also removes them when on Unix (or elsewhere). Even better, it removes trailing whitespace. Trailing whitespace should never be "significant", so removing it before processing a line is a good general practice.

- tye        

My perlport says

With default settings that function looks for a trailing "\n" character and thus trims in a portable way.

Did it use to say something different?

Thanks all for the ideas!

@Marshall: I'm not working on something specific, just playing around.

The one thing I like about chop is that its like pop, but for scalars.

The s/\s+$// regex which tye mentioned is pretty cool. Also, I like the ability of Perl to increment strings, as ikegami said. Bummer I didn't know that yet.

As for the padding, I think I'll stick with my old technique, e.g.: substr(("0"x9)."64",-4). I think its pretty self-explaining.

Thanks, Iwein