[Haskell-cafe] How does one create an input handle bound to a string instead of a file?

Bob Ippolito bob at redivi.com
Thu Feb 28 04:05:24 CET 2013


I haven't had time to make an example yet but it looks like if you go down
to GHC.IO.Handle.Internals there's a mkHandle function that takes a
BufferedIO and some other stuff and gives you an IO Handle.


On Wed, Feb 27, 2013 at 3:23 PM, Gregory Collins <greg at gregorycollins.net>wrote:

> Hm, perhaps I stand corrected. Then how exactly do you make the bytestring
> Handle?
>
>
> On Thu, Feb 28, 2013 at 12:15 AM, Don Stewart <dons00 at gmail.com> wrote:
>
>> I don't think that's right - Simon's buffer class rewrite should have
>> made this possible, I think.
>>
>>
>> http://hackage.haskell.org/packages/archive/base/4.2.0.1/doc/html/GHC-IO-BufferedIO.html
>> On Feb 27, 2013 10:52 PM, "Gregory Collins" <greg at gregorycollins.net>
>> wrote:
>>
>>> On Wed, Feb 27, 2013 at 9:38 PM, John D. Ramsdell <ramsdell0 at gmail.com>wrote:
>>>
>>>> How does one create a value of type System.IO.Handle for reading that
>>>> takes its input from a string instead of a file?  I'm looking for the
>>>> equivalent of java.io.StringReader in Java.  Thanks in advance.
>>>>
>>>
>>> You can't. There are several libraries that purport to provide better
>>> interfaces for doing IO in Haskell, like conduit, pipes, enumerator, and my
>>> own io-streams library (http://github.com/snapframework/io-streams,
>>> soon to be released). You could try one of those.
>>>
>>> G
>>> --
>>> Gregory Collins <greg at gregorycollins.net>
>>>
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>
>
> --
> Gregory Collins <greg at gregorycollins.net>
>
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