On Thu, Oct 20, 2005 at 09:59:49AM -0600, Luke Palmer wrote:
: How about this:
: 
:     sub foo(c|T $x) {
:         my sub util (c|T $in) {...}
:         util($x)
:     }
: 
: Is that c|T in util() a new, free type variable, or am I asserting
: that the type of util()'s argument must be the same type as $x?

It's a new T according to the current thinking.  Just use T if you
want the same one.  (But that does force util to be recloned on every
entry to foo, I expect.)

Larry

Reply via email to