[Haskell-cafe] Interpreter with Cont
ben.franksen at online.de
Mon Nov 21 22:57:37 CET 2011
You'll probably get answers from people who are more proficient with this,
but here's what I learned over the years.
Tim Baumgartner wrote:
> Is Cont free as well?
No. In fact, free monads are quite a special case, many monads are not free,
e.g. the list monad. I believe what David Menendez said was meant to mean
'modulo some equivalence relation' i.e. you can define/implement many monads
as 'quotients' of a free monad. But you cannot do this with Cont (though I
am not able to give a proof).
> I guess so because I heard it's sometimes called the
> mother of all monads.
It is, in the sense that you can implement all monads in terms of Cont.
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