Tony Morris

@djspiewak @nuttycom @jdegoes @milessabin Tell me what haskell gets wrong in this regard. Do not fail.

@dibblego @nuttycom @jdegoes @milessabin I’ve already listed a trivial example which proves the distinction is arbitrary.

@djspiewak @nuttycom @jdegoes @milessabin No, you really haven't. And worse, not even close.

@dibblego @nuttycom @jdegoes @milessabin Yes, I really have. Do you want me to instead show a type that has more than one valid Monad?

@djspiewak @nuttycom @jdegoes @milessabin Yes, because I can't think of one and that will totally demonstrate your point!

@dibblego @nuttycom @jdegoes @milessabin An utterly useless example of a type with two (both valid!) monads: http://pastebin.com/C6WX8kkP 

@djspiewak @nuttycom @jdegoes @milessabin Yes thanks. Note demonstrate the connection to your conclusion. No failing.

@dibblego @nuttycom @jdegoes @milessabin My conclusion is far less broad than you think it is.