in reply to Re^3: my $1
in thread my $1

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

-Mark

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Re^5: my $1
by ikegami (Patriarch) on Apr 24, 2009 at 02:30 UTC

    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.