Re: [Haskell] What's up with this Haskell runtime error message:

[Garrett Mitchener]

I came up with a system of coloring -- you'll have to view this message as html to see it.  You start with the input parameters -- those are green.  Anything defined in terms of green is blue.  Anything defined in terms of green & blue is purple.  Anything defined in terms of green, blue, and purple is orange, etc.  Then you can see the problem.  So in foo, everything checks out, you get to orange and there's just a couple of expressions left and you can see that there's no loop.

-----------------------------------------------------------------------------------------------------------


 foo  (step,r0,mu0) = bar (
step,r1,r0,mu1,mu0)
    where
    r1 = r0-
step*rd
    mu1 = mu0-step*mud


    rd = c*c*mu0
    mud
 = c*c/r0 - (foobar_r z)/c
    
c = baz(z)
    z = 6.378388e6-r0

baz z | z<125 = -0.25*z+1537.5

        | otherwise = 0.0169*z+1504.1

foobar_r z | z<125 = 0.25

    | otherwise = -0.0169

But when you try coloring bar, you get this far and then we're stuck.  The problem is clear: r depends on rdc, which depends on rd0, which depends on c0, which depends on z0, which depends on r.  So your definition for r includes a loop.

bar 
(step,r2,r1,mu2,mu1) = (r,z0) : bar (step,r1,r,mu1,m)
    where
    r = r2
+2*step*rdc
    m = mu2+2*step*mudc
    rdc = (
rd2+rd1+rd0)/6
    mudc = (mud2+mud1+mud0)/6


    rd2 = c2*c2*mu2
    mud2
 = c2*c2/r2 - (foobar_r z2)/c2


    rd1 = c1*c1*mu1
    mud1 = 
c1*c1/r1 - (foobar_r z1)/c1

    rd0 = c0*c0*m
    mud0 = c0*c0/r - (foobar_r z0)/c0


    c2 = baz(z2)

    c1 = baz(
z1)
    c0 = baz(z0)

    z2 = 6.378388e6-r2
    z1 = 6.378388e6
-r1
    z0 = 6.378388e6-r

main :: IO ()
main = do
    print $ take 100 (foo (0.1, 6.378388e6,0))

 

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