[Haskell-beginners] Why does floor not consider its argument an Integral?

[Jeffrey Brown]

I solved it. I was misreading the error report as being about the input to
floor, when the problem was the output of it. This code works:

splitArcAtIntegers:: Arc -> [Arc]
splitArcAtIntegers (a,b) = let
    ceiling_ish = fromInteger $ (floor a) + 1
  in if      b <= a           then []
     else if b <= ceiling_ish then [(a,b)]
     else (a,ceiling_ish) : splitArcAtIntegers (ceiling_ish,b)


On Sat, Nov 7, 2015 at 10:14 PM, Jeffrey Brown <jeffbrown.the at gmail.com>
wrote:

> Tidal [1] defines these data types:
>
>     type Time = Rational
>     type Arc = (Time, Time)
>
> I want to write a function "splitMultiCycArc" which divides an Arc into
> mostly-integer segments, so that, for instance,
>
>     splitMultiCycArc (0,1)   = [(0,1)]
>     splitMultiCycArc (0,2)   = [(0,1),(1,2)]
>     splitMultiCycArc (0,3)   = [(0,1),(1,2),(2,3)]
>     splitMultiCycArc (1%2,2) = [(1%2,1),(1,2)]
>     splitMultiCycArc (1,5%2) = [(1,2),(2,5%2)]
>
> I thought I had solved the problem with this code:
>
>     splitMultiCycArc:: Arc -> [Arc]
>     splitMultiCycArc (a,b) = let ceiling_ish = floor a + 1 in
>       if      b <= a           then []
>       else if b <= ceiling_ish then [(a,b)]
>       else (a,ceiling_ish) : splitMultiCycArc (ceiling_ish,b)
>
> When I try to load that, I get this single error:
>
>     > :reload
>     [12 of 13] Compiling Sound.Tidal.JBB  ( Sound/Tidal/JBB.hs,
> interpreted )
>
>     Sound/Tidal/JBB.hs:16:44:
>         No instance for (Integral Time) arising from a use of ‘floor’
>         In the first argument of ‘(+)’, namely ‘floor a’
>         In the expression: floor a + 1
>         In an equation for ‘ceiling_ish’: ceiling_ish = floor a + 1
>     Failed, modules loaded: Sound.Tidal.Strategies, Sound.Tidal.Dirt,
> Sound.Tidal.Pattern, Sound.Tidal.Stream, Sound.Tidal.Parse,
> Sound.Tidal.Tempo, Sound.Tidal.Time, Sound.Tidal.Utils,
> Sound.Tidal.SuperCollider, Sound.Tidal.Params, Sound.Tidal.Transition.
>     >
>
> And yet under other conditions, "floor" is perfectly happy operating on a
> Time value:
>
>     > let a = (1%2,1) :: Arc
>     > floor (fst a) + 1
>     1
>     >
>
> [1] https://hackage.haskell.org/package/tidal
>
>
> --
> Jeffrey Benjamin Brown
>



-- 
Jeffrey Benjamin Brown
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