2008/8/6 Galchin, Vasili <[EMAIL PROTECTED]>
> Hi Jeremy and the haskell community,
>
> I think now my question is why isn't there a higher-order "with"
> function in the Haskell Prelude?
>
There is. Well, not in the prelude, but in the base package at least. It's
called "bracket". But I think the reason that not all "with" functions are
implemented with bracket is because "with" is not as well defined as a
function; it is an idiom, and varies in too many ways to have a well-defined
semantics.
However, you could presumably abstract this more using typeclasses with
fundeps or associated types.
class Resource a where
type Constr a
acquire :: Constr a -> IO a
release :: a -> IO ()
with :: (Resource a) => Constr a -> (a -> IO b) -> IO b
with constr = bracket (acquire constr) release
I haven't calculated how many of the existing with* functions would fit into
this abstraction.
Luke
> Regards, Vasili
>
> On Wed, Aug 6, 2008 at 2:25 AM, Jeremy Shaw <[EMAIL PROTECTED]> wrote:
>
>> At Wed, 6 Aug 2008 00:43:55 -0500,
>> Galchin, Vasili wrote:
>> >
>> > [1 <multipart/alternative (7bit)>]
>> > [1.1 <text/plain; ISO-8859-1 (7bit)>]
>> > Hello,
>> >
>> > 1) Is there a common assumed semantics across all of the Haskell
>> "with"
>> > things? withString? withData?
>>
>> A vague semantic is that some resource is acquired, used, and then
>> released.
>>
>> > 2) If the answer to 1) is yes, then does this imply some kind of
>> > polymorphism? Couldn't "with" be "reduced" to a higher order function
>> then
>> > ??
>>
>> I think Control.Exception.bracket is used to implement many, but not
>> all, of the with* functions.
>>
>> > newThing :: IO a
>> > newThing = ...
>>
>> > destroyThing :: a -> IO b
>> > destroyThing ...
>>
>> > withThing :: (Thing -> IO a) -> IO a
>> > withThing useThing = bracket newThing destroyThing useThing
>>
>> I have not fully answered either of your questions, but hopefully this
>> is a start.
>>
>> j.
>>
>
>
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