View source for Talk:Correctness of short cut fusion

If unfoldr would use a lazy pattern match:
<haskell>
unfoldr :: (b -> Maybe (a,b)) -> b -> [a]
unfoldr p e = case p e of Nothing     -> []
                          Just ~(x,e') -> x:unfoldr p e'
</haskell>
the left hand side of the example without seq will be the same as the right hand side:
<haskell>
destroy g (unfoldr p e) = g step (unfoldr p e)
                        = case step (unfoldr p e) of Just z -> 0
                        = case step (case p e of Nothing     -> []
                                                 Just ~(x,e') -> x:unfoldr p e') of Just z -> 0
                        = case step (case Just undefined of Nothing     -> []
                                                            Just ~(x,e') -> x:unfoldr p e') of Just z -> 0
                        = case step (undefined:unfoldr p undefined) of Just z -> 0
                        = case Just (undefined,unfoldr p undefined) of Just z -> 0
                        = 0
</haskell>
--[[User:Twanvl|Twanvl]] 12:18, 13 February 2007 (UTC)

== Details about what uses of seq are dangerous ==

If I understand things properly, the essential problem with <hask>seq</hask> in <hask>foldr/build</hask> is that it allows the builder function
<haskell>
g :: forall b . (a -> b -> b) -> b -> b
g cons nil = ...
</haskell> 
to do something with values of type <hask>b</hask> other than pass them to <hask>cons</hask>, namely to <hask>seq</hask> against them. If <hask>g</hask> doesn't force anything whose type includes <hask>b</hask>, it should, I believe, be safe to fuse. For example, <hask>unfoldr</hask> written using <hask>build</hask> should, I believe, be safe to fuse because the function it is given isn't passed the polymorphic cons and nil arguments&mdash;the list generation is left up to the known-<hask>seq</hask>-free machinery of <hask>unfoldr</hask>. --[[User:Dfeuer|Dfeuer]] ([[User talk:Dfeuer|talk]]) 03:22, 15 August 2014 (UTC)