[Haskell-cafe] Monad explanation

Lennart Augustsson lennart at augustsson.net
Mon Feb 9 06:39:44 EST 2009

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 <dev at mobileink.com>:
> On Mon, Feb 9, 2009 at 4:38 AM, Tony Morris <tmorris at tmorris.net> wrote:
>> Hash: SHA1
>> 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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