Re: [stack] What does "concatenative" actually mean?

Haskell *does* need a monad to do I/O.  The essential
difference between "read" having a stack shape of 
F -> F O, and Haskell's IO monad, is that the Haskell
IO monad is guaranteed to confine any side effects to
the IO monad; that is, the result of an IO operation will
always have a type of the form IO <something>.  And 
there's no way any result depending on an I/O operation
can avoid having type IO <something>.  There's no
escape from the IO monad in Haskell.  Lacking a static
type system, Joy offers no such guarantee.

Cat, OTOH, has a Hindley-Milner static type system.
Perhaps Cat could do monads.

-Rod








William Tanksley scripsit:

> The truly pure functional languages, like Haskell, have a mathematical
> model for state-changing effects, like objects and I/O. The problem is
> that purely functional languages don't express what order their
> operations will occur in; they just tell the computer what operations to
> do, and let the compiler figure out the best order. So Haskell provides
> a mathematical construct called a "monad". Monads can mathematically
> only be evaluated in order, so the compiler receives enough information
> to know what order to execute your code in.

You don't really need monads: they are just a convenience, because
it would be painful to thread state through every function in
your program just in case a callee wants to use some of it; see
http://lambda-the-ultimate.org/node/1155#comment-12726 for details.
In Joy, the equivalent of the I/O monad is the simple fact that "read"
has a stack shape of F -> F O, preserving the F object on the stack.

-- 
One Word to write them all, John Cowan <[email protected]>
One Access to find them, http://www.ccil.org/~cowan
One Excel to count them all,
And thus to Windows bind them. --Mike Champion





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