[Haskell-beginners] How best to do this?

[David McBride]

Couldn't it just use fromEnum and compare the integers you get and
figure out which is bigger?

On Mon, Apr 29, 2013 at 2:17 PM, Brent Yorgey <byorgey at seas.upenn.edu> wrote:
> But then \x y -> [x .. y]  would have to have the type
>
>   (Ord a, Enum a) => [a]
>
> whereas now it just has the Enum constraint.  Either that or else the
> notation would work differently for literals vs. expressions but that
> would be just awful.
>
> -Brent
>
> On Mon, Apr 29, 2013 at 11:57:38AM -0400, David McBride wrote:
>> I think it is because that syntax is desugared directly to enumFromTo
>> x y and enumFromTo 3 0 is [].  Random things would probably blow up if
>> you make that function work in reverse.  But I would love it if ghc
>> just checked whether the first was lower than the second and swapped
>> them as a convenience.
>>
>> On Mon, Apr 29, 2013 at 11:44 AM, emacstheviking <objitsu at gmail.com> wrote:
>> > ROFLCOPTER indeed Batman!
>> >
>> > I had no idea of that... I just *assumed* (usual rules apply I guess) that
>> > [3..0] was the "opposite" of [0..3] but sure enough a wuicj ghci session
>> > reveals the bitter truth!
>> >
>> > Thanks again... i can see that it's not just me that is too lazy at times. I
>> > guess writing [3,2..0] will do for now but is that a bug or is there some
>> > other reasoning behind it?
>> >
>> > We live and learn, well, I live anyway...
>> >
>> > :)
>> >
>> >
>> >
>> > On 29 April 2013 16:37, David McBride <toad3k at gmail.com> wrote:
>> >>
>> >> One other gotcha.  I don't know why it is this way, but [3..0]
>> >> evaluates to [].  I have no idea why reverse notation is not allowed.
>> >> But you can just manually reverse it or you can go [3,2..0].
>> >>
>> >> On Mon, Apr 29, 2013 at 11:26 AM, emacstheviking <objitsu at gmail.com>
>> >> wrote:
>> >> > damn that lazy evaluation! LMAO  ...a good point brent and yuo have no
>> >> > doubt
>> >> > saved me hours of head scratching this evening when I try out the "new
>> >> > improved software". Oh dear oh dear oh dear...
>> >> >
>> >> > doOption dev (Forward n)  = do
>> >> >   putStrLn $ "> STEP FORWARD " ++ (show n)
>> >> >   stepBits dev ioPORTA [3..0]
>> >> >
>> >> >
>> >> > doOption dev (Backward n) = do
>> >> >   putStrLn $ "> STEP BACKWARD " ++ (show n)
>> >> >   stepBits dev ioPORTA [0..3]
>> >> >
>> >> > stepBits dev port = mapM_ stepIt
>> >> >   where stepIt bit = mapM_ (\s -> HW.setPortBit dev port bit s >>
>> >> > stepDelay)
>> >> > [0,1]
>> >> >
>> >> > I now have the above as my current "final" implementation... hopefully
>> >> > that
>> >> > *does* do what I think it does because mapM_ is driving it and will
>> >> > cause
>> >> > evaluation of the actions?
>> >> >
>> >> >
>> >> >
>> >> >
>> >> > On 29 April 2013 15:56, Brent Yorgey <byorgey at seas.upenn.edu> wrote:
>> >> >>
>> >> >> On Mon, Apr 29, 2013 at 02:59:29PM +0100, emacstheviking wrote:
>> >> >> > I have built a library for using the Hexwax expandIO-USB chip and I
>> >> >> > have
>> >> >> > now got some code to drive a stepper motor:
>> >> >> >
>> >> >> > doOption :: HWHandle -> Flag -> IO ()
>> >> >> > doOption dev (Backward n) = do
>> >> >> >   putStrLn $ "> STEP BACKWARD " ++ (show n)
>> >> >> >   let x = [ stepBit b | b <- [3..0]]
>> >> >> >   return ()
>> >> >> >     where
>> >> >> >       stepBit p b = setBit p b 0 >> setBit p b 1
>> >> >> >         where setBit p b s = HW.setPortBit dev p b s >> stepDelay
>> >> >>
>> >> >> The other posted solutions are good, but I also want to make a very
>> >> >> important comment about the above code: it does not actually step any
>> >> >> bits!  All it does is print some stuff.  x is simply a name for a list
>> >> >> of IO actions; it is never used so it just gets garbage collected and
>> >> >> the IO actions are never run.
>> >> >>
>> >> >> -Brent
>> >> >>
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