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

[Jeffrey Brown]

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
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