[Haskell-cafe] Interpreter with Cont

Ben Franksen 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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