Re: [Haskell-cafe] Monad explanation

Huh? The getChar function does yield a value of type (IO Char), exactly as the type signature says. If you want access to the Char you must use a >>=, just like in any other monad. 2009/2/9 Gregg Reynolds :

On Mon, Feb 9, 2009 at 4:38 AM, Tony Morris wrote:

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I also agree it is a value. The original post was attempting to make a distinction that does not exist. I deliberately avoided that topic.

"A thing cannot be both a value and a function, but e,g, getChar"

My original intent was to hope the poster reconsidered the whole post. You've blown my cover :)

My bad, I restate: a value cannot be both static and dynamic. Or an object and a morphism. Or an element and a function. Sure, you can treat a morphism as an object, but only by moving to a higher (or different) level of abstraction. That doesn't erase the difference between object and morphism. If you do erase that difference you end up with mush. getChar /looks/ like an object, but semantically it must be a morphism. But it can't be a function, since it is non-deterministic. So actually the logical contradiction comes from the nature of the beast.

Another reason it's confusing to newcomers: it's typed as "IO Char", which looks like a type constructor. One would expect getChar to yield a value of type IO Char, no? But it delivers a Char instead. This is way confusing. So I take "type IO foo" to mean "type foo, after a side effect". In a sense "getChar :: IO Char" isn't even a true type signature.

In any case, many thanks to all who have contributed to the thread. It's sharpened my thinking revealed weaknesses in my terminology, and I expect I'll make my inevitable contribution to the infinite Haskell tutorial on the topic before too long.

-gregg

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