Can we deconstruct arbitrary type constructors?

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In Haskell, or any other programming language supporting pattern matching, can we achieve the following:

Given a type constructor e.g. Left :: a -> Either a b I can extract its argument with

let f (Left x) = [x]

Is there any way to achieve the same thing for any type constructor, something like an operator (*) to lift Left x to Apply (λa.Left a) [x] or BinTree x y to Apply (λa,b.BinTree a b) [x,y], so we could do

let extract x = case *x of Apply _ xs -> xs

asked 33 secs ago

Adam

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Can we deconstruct arbitrary type constructors?