# Talk:Correctness of short cut fusion

If unfoldr would use a lazy pattern match:

```unfoldr :: (b -> Maybe (a,b)) -> b -> [a]
unfoldr p e = case p e of Nothing     -> []
Just ~(x,e') -> x:unfoldr p e'
```

the left hand side of the example without seq will be the same as the right hand side:

```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
```

--Twanvl 12:18, 13 February 2007 (UTC)

## Details about what uses of seq are dangerous

If I understand things properly, the essential problem with `seq` in `foldr/build` is that it allows the builder function

```g :: forall b . (a -> b -> b) -> b -> b
g cons nil = ...
```

to do something with values of type `b` other than pass them to `cons`, namely to `seq` against them. If `g` doesn't force anything whose type includes `b`, it should, I believe, be safe to fuse. For example, `unfoldr` written using `build` should, I believe, be safe to fuse because the function it is given isn't passed the polymorphic cons and nil arguments—the list generation is left up to the known-`seq`-free machinery of `unfoldr`. --Dfeuer (talk) 03:22, 15 August 2014 (UTC)