On 31 Aug 2002 at 0:17, Piers Cawley wrote:
> my $pattern = rx:w / $1:=(\S+) =<gt> $2:=(\S+) |
> $2:=(\S+) <lt>= $1:=(\S+) /;
>
> @ary = m/<$pattern>/
>
> how many elements are there in @ary? I can
> make a case for 4 quite happily. Certainly that's what A5 seems to
> imply:
>
> Suppose you want to use a hypothetical variable to bind a name to
> a capture:
>
> / (\S+) { let $x := $1 } /
>
> A shorthand for that is:
>
> / $x:=(\S+) /
>
> The parens are number independently of any name, so $x is an alias
> for $1.
>
> And it's that last sentence that's important here. So, it looks like
> C<< +@ary >> is going to be 4.
How could it be 4? If the example would've been
> my $pattern = rx:w / $a:=(\S+) =<gt> $b:=(\S+) |
> $b:=(\S+) <lt>= $a:=(\S+) /;
Then there is 4 variables to speak of ($1,$2,$a,$b) and a question
arises about which of these are returned.
In the original example however we only have 2 variables ($1,$2) so
it can't really return anything else than those 2.
> m: w / $2:=(\S+) <lt>= $1:=(\S+) /
>
> Now, assignment to hypotheticals doesn't happen all at once, it
> happens when the matching engine reaches that point in the
> string. Unless I'm very much mistaken, the order of execution will
> look like:
>
> $2:=$1; $1:=$2;
>
> And it seems to me that we'll end up with $1 and $2 both bound to the
> same string; which isn't working quite how I expect. Or do we special
> case the occasions when we bind the result of a capturing group to a
> numeric match variable?
As I understand it, binding to $1 etc.. is a special case. Also I
don't see any problems in your example:
m: w / $2:=(\S+) <lt>= $1:=(\S+) /
First () is captured and assigned to $2 (instead of $1).
Then second () is captured and assigned to $1.
--
Markus Laire 'malaire' <[EMAIL PROTECTED]>
