On Monday, 18. August 2008 20:38:05 Patrick R. Michaud wrote: > I would somewhat expect > a reference to be instead handled using a statement like > > $foo[1] := $bar; > > Comments and clarifications appreciated.
I would also opt for copy semantics whenever = is used for assignment. But it seems to be the case that this is not deep, just like captures are only one level deep readonly. So, I would also expect $foo[1] = \$bar to result in 24. Regards, TSa. -- "The unavoidable price of reliability is simplicity" -- C.A.R. Hoare "Simplicity does not precede complexity, but follows it." -- A.J. Perlis 1 + 2 + 3 + 4 + ... = -1/12 -- Srinivasa Ramanujan
