Re: [Haskell-cafe] Interpreter with Cont

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. Cheers Ben