Re: [Haskell-cafe] ($) and ApplicativeDo

Wow, that's pretty unexpected. Another repro case: p = pure bar :: Applicative f => f a bar = do x <- pure "ok" p x ... gives the same error. So ApplicativeDo statements *must* end with exactly the statement "pure"?

El 16 jun 2016, a las 12:39, Joe Quinn escribió:

I expect it's that the root of the expression on the last line is ($) and not pure. ApplicativeDo has restrictions on what you can do to avoid allowing blocks that have to translate in terms of join/(>>=), and likely one of them is that the last line can't be an arbitrary thing of the right type.

...

On 6/16/2016 12:30 PM, amindfv@gmail.com wrote: They're the same:

...

:t \x -> pure x \x -> pure x :: Applicative f => a -> f a :t \x -> pure $ x \x -> pure $ x :: Applicative f => a -> f a

Tom

El 16 jun 2016, a las 12:24, KC escribió:

...

Think of the types of

pure x

And

pure $ x

-- --

Sent from an expensive device which will be obsolete in a few months! :D

Casey

...

On Jun 16, 2016 9:06 AM, wrote: foo :: Applicative f => f String foo = do x <- pure "this works" pure x

... but replace "pure x" with "pure $ x" and it doesn't typecheck: a monad instance is required!

Tom _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://mail.haskell.org/cgi-bin/mailman/listinfo/haskell-cafe

_______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://mail.haskell.org/cgi-bin/mailman/listinfo/haskell-cafe

_______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://mail.haskell.org/cgi-bin/mailman/listinfo/haskell-cafe