[Haskell-cafe] ANNOUNCE: pipes-core 0.1.0

Ben Franksen ben.franksen at online.de
Tue Apr 17 23:23:37 CEST 2012

Paolo Capriotti wrote:
> On Mon, Apr 16, 2012 at 10:13 PM, Ben Franksen <ben.franksen at online.de>
> wrote:
>> (1) What is the reason for the asymmetry in
>> type Producer b m = Pipe () b m
>> type Consumer a m = Pipe a Void m
>> i.e. why does Producer use () for the input? I would expect it to use
>> Void, like Consumer does for its output. Calling await in a Producer
>> resulting in an immediate 'return ()' as you say is allowed (in the
>> tutorial) strikes me as not very useful.
> The underlying reason for the asymmetry is the fact that '()' is a
> terminal object in the category of haskell types and *total*
> functions, while 'Void' is an initial object.
> Here's a property that uniquely determines the definitions of
> 'Producer' above. Let 'X' be the type such that 'Producer b m = Pipe X
> b m'. For all producers 'p' there should be a unique (total) pipe
> 'alpha :: forall a r. Pipe a X m r' such that 'alpha >+> p' and 'p'
> are observationally equal. In other words, since a producer "never
> uses values of its input type 'a'", there should be a unique way to
> make it into a pipe which is polymorphic in 'a'. It's easy to see that
> this property immediately implies that 'X' should be a terminal
> object, i.e. '()', and 'alpha' is therefore 'pipe (const ())'.
> Dually, you obtain that 'Consumer a m' is necessarily 'Pipe a Void m',
> and 'alpha = pipe absurd'.

Ok, thanks for the explanation. Makes sense...

>> (2) The $$ operator is poorly named. I would intuitively expect an
>> operator that looks so similar to the standard $ to have the same
>> direction of data flow (i.e. right to left, like function application and
>> composition) but your is left to right. You could use e.g. >$> instead,
>> which has the additional advantage of allowing a symmetric variant for
>> the other direction i.e. <$<.
> '$$' is inspired by iteratees. Similarly to its iteratee counterpart,
> it discards upstream result values and only returns the output of the
> last pipe. That said, '>$>' looks like a clearer alternative, so I
> could consider changing it.

(...or maybe use a plain function instead of an operator...)


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