In another thread of mine someone responded with a snippet that I've been trying to wrap my head around ever since. This subject is a major tangent from the other thread, so I think it should be on its own. I hope that's OK.

Here is the original snippet. It is the contents of the s// regexp operator that has me completely stumped.

%d = (
    foo => sub { "[@_]"; },
    bar => sub { "(@_)"; },
);

$_ = 'foo fooz bar baz1';
s/(\w+)(?(?{ !$d{$1} })(?!))/$d{$1}->($1)/ge;
print "$_\n";

[download]

Here is the output.

[foo] [foo]z (bar) baz1

[download]

I added comments to the code in my attempt to understand what was happening. Most of my explanatory text is contained therein.

s/
    (\w+)   # One or more word chars in capture buffer 1. Got it.
    (   # Embedding a regexp modification character or what?
        # So everything within this set of parenthesis is
        # supposed to evaluate to either m, s, i, x, p, g, or c?
        # I don't get it.
        # Also, why is all this on the left side of s\\? Why are we
        # _matching_ this?
        ?(?{ !$d{$1} }) # This is really just used as a boolean to
                        # indicate that a hash key exists. I get it,
                        # kind of.
        (?!)    # From perldoc, "a zero-width negative look-ahead
                # assertion. I might understand this if only there was
                # a pattern after the "?!".
    )
/$d{$1}->($1)/gex;
print "$_\n";

# Here is my walkthrough for 'foo'.
# 1. (\w+) matches 'foo'.
# 2. $d{$1} is _true_. !_true_ = false.  In this case, false is "1".
# 3. (?!) says to *not* match a "1" followed by null. Maybe?
# 4. (?...) means use the 1 not followed by null as a regexp modifier
#    to 'foo'? What?!

[download]

A few of my big failures to comprehend I think are:

  1. My thinking that the outer set of *(?...)* is the use of a regexp mod. But if it's not that, then what is it?
  2. The *(?!)* without a pattern following it. Does he really want to match any '1' without a following null?
  3. Why are these on the left side of the s//? Why are we matching something other than the word?

Thanks.

EDIT: I updated the $_ var to include the "baz1" string.

In reply to Regular expression operators (?{code}), (?!pattern), and (?...) by jffry

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