Am Mittwoch 16 Dezember 2009 05:47:20 schrieb Gregory Crosswhite:
On Dec 15, 2009, at 8:28 PM, Daniel Fischer wrote:
Am Mittwoch 16 Dezember 2009 05:08:39 schrieb Gregory Crosswhite:
Not even then, necessarily. And it's not always a good idea.
f k = [1 .. 20^k]
You raise a really good point here. One can force sharing, as I understand it, by using a let clause:
n = let xs = f 20 in length (xs ++ xs)
If I understand correctly, this should cause xs to be first evaluated,
It is evaluated during the calculation of the length. But whereas in n = let xs = f 20 ys = f 10 in length (xs ++ ys) every cell of xs can be immediately garbage collected - so it will only take insanely long to get a result -, in the above example the scond argument of (++) holds on to xs, so they can't be garbage collected (I wouldn't bet any body parts on that, an implementation is allowed to recalculate a named entity on every occurrence, but it's the expected behaviour) and you will run into memory problems. The good part of it is that it finishes much faster :)
and then cached until the full length is computed, which in this case is obviously undesirable behavior.
Cheers, Greg