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Re^3: my $1

by ikegami (Pope)
on Apr 23, 2009 at 21:50 UTC ( #759678=note: print w/replies, xml ) Need Help??


in reply to Re^2: my $1
in thread my $1

You're right. I still had this on the mind. It was posted only two hours before.

Back to your problem. There isn't one if you don't use globals. Fix:

if (my ($pre, $post) = m/(.+?)\+(.+)/) { return diff($pre)."+".diff($post); }

Replies are listed 'Best First'.
Re^4: my $1
by mltucker (Initiate) on Apr 24, 2009 at 01:30 UTC

    Nice solution. Thanks! Time to do some perldoc sifting to sort this all out :-).

    -Mark

      Thanks. My earlier code explained:

      • my ($pre, $post) is short for (my $pre, my $post).

      • A list-like expression on the LHS of = (such as my ($pre, $post)) causes the list assignment (aassign) operator to be used instead of a scalar assignment (sassign) operator.

      • The list assignment operator evaluates its RHS operand in list context.

      • In list context, m// (no "g") returns

        • () if the match failed,
        • the captures if the match succeeded and the pattern contains captures, or
        • 1 (one) if the match succeeded and the pattern contains no captures.
      • The list assignment operator returns the number of elements returned by its RHS operand.

      • The if condition will be false when the match fails because the list assignment operator returned 0 (zero) because the match operator returned zero elements.

      • The if condition will be true when the match succeeds because the list assignment operator returned 2 (two) because the match operator returned two captures.

      • And finally, each level of recursion of the function get its own pad, which means my ($pre, $post) creates new variables for each level of recursion, which means the they keep their value even through a recursive call.

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