Hey guys,
Right now I'm facing with a type problem that is really nasty, I want to compose a list of enumeratees using the ($=) operator to create a new enumerator. Whenever I'm trying to use the foldx function in conjunction with ($=) I get this error:
> :t foldr ($=)
<interactive>:1:7:
Occurs check: cannot construct the infinite type:
b0 = Step ao0 m0 b0
Expected type: Enumerator ao0 m0 (Step ao0 m0 b0)
-> Enumeratee ao0 ao0 m0 b0
-> Enumeratee ao0 ao0 m0 b0
Actual type: Enumerator ao0 m0 (Step ao0 m0 b0)
-> Enumeratee ao0 ao0 m0 b0
-> Enumerator ao0 m0 b0
In the first argument of `foldr', namely `($=)'
In the expression: foldr ($=)
> :t Prelude.foldl ($=)
<interactive>:1:15:
Occurs check: cannot construct the infinite type:
b0 = Step ao0 m0 b0
Expected type: Enumerator ao0 m0 (Step ao0 m0 b0)
-> Enumeratee ao0 ao0 m0 b0
-> Enumerator ao0 m0 (Step ao0 m0 b0)
Actual type: Enumerator ao0 m0 (Step ao0 m0 b0)
-> Enumeratee ao0 ao0 m0 b0
-> Enumerator ao0 m0 b0
In the first argument of `Prelude.foldl', namely `($=)'
In the expression: Prelude.foldl ($=)
<interactive>:1:15:
Occurs check: cannot construct the infinite type:
b0 = Step ao0 m0 b0
Expected type: Enumerator ao0 m0 (Step ao0 m0 b0)
-> Enumeratee ao0 ao0 m0 b0
-> Enumerator ao0 m0 (Step ao0 m0 b0)
Actual type: Enumerator ao0 m0 (Step ao0 m0 b0)
-> Enumeratee ao0 ao0 m0 b0
-> Enumerator ao0 m0 b0
In the first argument of `Prelude.foldl', namely `($=)'
In the expression: Prelude.foldl ($=)
Obviously there is something I don't quite understand about the ($=) (=$) functions, how can one compose a list of enumeratees, is it even possible?
Cheers.
Roman.-