Nested dummy procedures

Hello,

Does any body know how make code below more smart? Do not look whta
functions do -- it's example.
Code:
module Parameters
implicit none

integer, parameter :: K = 10, L = 10

end module Parameters

Code:
program DummyProcedures
use Parameters

implicit none

real(8) X(L)

X = Newton( MyFunc, X )

contains
function Newton( Func, Xi ) result(X)
real(8) :: X(L)
interface
function Func( V )
use Parameters
real(8), intent(in) :: V(L)
real(8) Func(K)
end
end interface
real(8), intent(in) :: Xi(L)

real(8) M(K, L)

M = DiffVectorByVector( Func, Xi )
X = M(1, :)
end function Newton

function DiffVectorByVector( Func, V ) result(M)
interface
function Func( V )
use Parameters
real(8), intent(in) :: V(L)
real(8) Func(K)
end
end interface
real(8), intent(in) :: V(L)
real(8) :: M(K, L)

M(1:K, 1) = Func( V )
end function DiffVectorByVector

function MyFunc( V )
real(8), intent(in) :: V(L)
real(8) MyFunc(K)

MyFunc = V(1:Min( K, L ))
end function MyFunc
end program DummyProcedures


I mean, when dummy function returns scalar all looking good.

Code:
program DummyScalarProcedure

implicit none

integer, parameter :: K = 10, L = 10
real(8) X(L)

X = Newton( MyFunc, X )

contains
function Newton( Func, Xi ) result(X)
real(8) :: X(L)
real(8) Func
real(8), intent(in) :: Xi(L)

real(8) M(K, L)

M = DiffVectorByVector( Func, Xi )
X = M(1, :)
end function Newton

function DiffVectorByVector( Func, V ) result(M)
real(8) Func
real(8), intent(in) :: V(L)
real(8) :: M(K, L)

M(K, 1) = Func( V )
end function DiffVectorByVector

function MyFunc( V )
real(8), intent(in) :: V(L)
real(8) MyFunc

MyFunc = V(Min( K, L ))
end function MyFunc
end program DummyScalarProcedure
It seems that when I declare dummy function as real(8) Func(K) compiler
thinks that it can be only array, but not a function which returns
array of length K! But real(8) Func -- can be function wich returns
scalar. So I had to use interfaces => module or triple declaration of K
and L. May be there is a way of writing more smart code in this case.
I'll be glad any help.

Stanislaus

.