GHC's use of a State-like monad to represent I/O is semantically bogus, a
fact that rears its head in a few compiler transformations (including
strictness analysis). I wouldn't take it as a good mental model.
On Sat, Sep 18, 2021, 7:09 PM Olaf Klinke
Perhaps one thing that speaks in favour of Anthony Clayden's suggestion to learn lambda calculus before/along Haskell is this: Haskell shares the evaluation model with the lambda calculus. One of Michael Turner's points of criticism was, as I understand it, insufficient documentation about the inner workings of Haskell [1]. Haskell needs constructs like the ST monad because of the evaluation model. Once the learner knows, perhaps by experimenting with it on paper or on [2,3], that a lambda term can be reduced in several ways, and confluence guarantees that all ways result in the same normal form, then some things might be much clearer: * Which freedoms the compiler can exercise in evaluating your program. * Why one needs the State monad for ordering sequences of effects (the programmable semicolon) [4]. * How the various tricks and pitfalls around lazy and strict functions work, e.g. Viktor's new Foldable documentation.
I looked at the first chapter of my K&R. It presents small example C programs and explains what they do. Taking small Haskell programs and explaining how these lambda terms are reduced to normal form would be a comparable (and desirable, in Michael's view) approach? The Haskell language report does not say anything in this respect, as far as I can see. Only translation of syntactic constructs to core Haskell is given.
Olaf
[1] Of course the documentation is out there somewhere, but not written in a style which pleases learners like Michael. [2] https://lambdacalc.io/ [3] https://capra.cs.cornell.edu/lambdalab/ [4] I tried to convince myself that the state monad indeed sequences effects. In a pure lambda world I think this should mean that a certain lambda term has only one way of being reduced. Is the following valid reasoning? type State s a = s -> (a,s) When Church-encoding the tuple (a,s) one finds that (m :: State s a) >>= (k :: a -> State s b) = \s -> m s k so the only applicable rule is beta-reducing m when applying to some s.
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