Question

I have been trying several regular expressions in the substitution operator:

$str =~ s/^0+(.)/$1/;

converts 0000 to 0 and 0001 to 1

$str =~ s/^0+./$1/;

converts 0000 to empty string, 000100 to 00, 0001100 to 100.

what difference is the parentheses making?

Answer 1

This seems like a bit of a misuse to me – you need the () to identify what’s your match.

http://perldoc.perl.org/perlre.html

Capture buffers

The bracketing construct ( … )
creates capture buffers. To refer to
the current contents of a buffer later
on, within the same pattern, use \1
for the first, \2 for the second, and
so on. Outside the match use “$”
instead of “\”. (The \ notation
works in certain circumstances outside
the match. See the warning below about
\1 vs $1 for details.) Referring back
to another part of the match is called
a backreference.

So basically you can use

if ($str =~ /^0+(.)/) { print "matched $1"; }

If you have more than one grouped matches they will be $1, $2, $3… etc e.g.

if ($str =~ /(0*)(1*)/) { print "I've got $1 and $2"; }