On 4/5/06, ihope <[EMAIL PROTECTED]> wrote:
I will be glad to. But just to make it more simple, it is a recursive function with a self referential set of definitions that builds a list like this :
------------------------------------------------------------------------------------------------------------
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
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))
On 4/5/06, Michael Goodrich <[EMAIL PROTECTED]> wrote:
> Looks like my calulation involves a self referential set of definitions.
>
> Is Haskell not able to deal with a self referential set of definitions?
Yes, it is, but not if that definition doesn't evaluate to a "proper"
value. For example:
main = do
print x
where x = 3 * x^2
What do you expect this to do?
It may help if you toss us the offending code.
I will be glad to. But just to make it more simple, it is a recursive function with a self referential set of definitions that builds a list like this :
------------------------------------------------------------------------------------------------------------
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
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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