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 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