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

Hmm, unless I have a different bug than I thought I did, this isn't working for me.

-Mark

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Re^3: my $1
by ikegami (Patriarch) on Apr 23, 2009 at 21:50 UTC

    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);
}

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

      -Mark
[reply]

        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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