Copying the list, sorry. I have a lot of trouble replying correctly with gmail's interface for some reason. :)
On Tue, Jan 10, 2012 at 11:14 AM, Dan Doel <dan.d...@gmail.com> wrote: > On Tue, Jan 10, 2012 at 5:01 AM, Simon Marlow <marlo...@gmail.com> wrote: >> On 09/01/2012 04:46, wren ng thornton wrote: >>> Shouldn't (# T #) be identical to T? >> >> No, because (# T #) is unlifted, whereas T is lifted. In operational terms, >> a function that returns (# T #) does not evaluate the T before returning it, >> but a function returning T does. This is used in GHC for example to fetch a >> value from an array without evaluating it, for example: >> >> indexArray :: Array# e -> Int# -> (# e #) I don't really understand this explanation. (# T #) being unlifted would mean it's isomorphic to T under the correspondence e <-> (# e #). _|_ = (# _|_ #) : (# T #), so this works. Does the difference have to do with unboxed types? For instance: foo :: () -> Int# foo _ = foo () bar :: () -> (# Int# #) bar _ = (# foo () #) baz = case bar () of _ -> 5 -- 5 quux = case foo () of _ -> 5 -- non-termination Because in that case, either (# Int# #) is lifted, or the Int# is effectively lifted when inside the unboxed tuple. The latter is a bit of an oddity. -- Dan _______________________________________________ Glasgow-haskell-users mailing list Glasgow-haskell-users@haskell.org http://www.haskell.org/mailman/listinfo/glasgow-haskell-users
