Re^3: 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);
}
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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	[reply]
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.	[reply] [d/l] [select]