Re: Re: Make InputStreamResource parameterless

> sorry for the late answer.
>
> ===
> def readXml(isr: InputStreamResoure[InputStream]) = ...
> ===
>
> Is too long (unnecessary complex).
>
> ===
> def readXml(isr: InputStreamResource) = ...
> ===
>
> is much easier to write.
>
> I'm proposing to make InputStreamResource work only with InputStream,
> not BufferedInputStream or any other InputStream.
>
> Currenly InputStreamResource is:
>
> ===
> abstract class InputStreamResource[+I <: InputStream] ...
> ===
>
> I'd like to make it:
>
> ===
> abstract class InputStreamResource ...
> ===
>
> S.
>
> On Fri, Nov 28, 2008 at 22:11, Jorge Ortiz <jorge.ortiz@...> wrote:
>> Also, what code do you find unnecessarily complex? Code in scalax or your
>> own code?
>>
>> On Fri, Nov 28, 2008 at 11:06 AM, Jorge Ortiz <jorge.ortiz@...> wrote:
>>>
>>> Those types are passed on to ManagedResource which uses them in a fair
>>> number of methods.
>>>
>>> I guess I'm not sure exactly what your proposing. Make a patch and e-mail
>>> it to the list before committing any major changes?
>>>
>>> --j
>>>
>>> On Fri, Nov 28, 2008 at 10:38 AM, Stepan Koltsov <yozh@...> wrote:
>>>>
>>>> OK, nobody against, I'll it :)
>>>>
>>>> S.
>>>>
>>>> On Mon, Nov 24, 2008 at 02:51, Stepan Koltsov <yozh@...> wrote:
>>>> > Hi, all.
>>>> >
>>>> > I'd like to make InputStreamResource and friends parameterless (i. e.
>>>> > WriterResource will work only with Writer).
>>>> >
>>>> > I find parametrization useless, and code using this class unnecessary
>>>> > complex.
>>>> >
>>>> > What do you think?
>