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.-
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