I don't think that you need nor that you should use generated functions.
But maybe I'm wrong, what are you trying to achieve? This should work
as you want:
function testfun2!{N}(X,Y::NTuple{N,Float64})
for i in eachindex(X), j in 1:N # much better to have the loop this way
X[i][j] = Y[j]
end
return X
end
# Setup for function call
InnerArrayPts = 3
OuterArrayPts = 10
Xinput = [Array{Float64}(InnerArrayPts) for r in 1:OuterArrayPts]
Yinput = rand(InnerArrayPts)
testfun2!(Xinput,tuple(Yinput...))
On Tue, 2015-09-29 at 13:20, Alan Crawford <a.r.crawf...@gmail.com> wrote:
> I would like to preallocate memory of an array of arrays and pass it to a
> function to be filled in. I have created an example below that illustrates
> my question(s).
>
> Based on my (probably incorrect) understanding that it would be desirable
> to fix the type in my function, I would like to be able to pass my array of
> arrays, X, in a type stable way. However, I can't seem to pass
> Array{Array{Float64,N},1}. If, however, i do not attempt to impose the type
> on the function, it works.
>
> Is there a way to pass Array{Array{Float64,N},1 to my function? Do I even
> need to fix the type in the function to get good performance?
>
> # version of Julia: 0.4.0-rc3
>
> @generated function
> testfun1!{N}(X::Array{Array{Float64,1},1},Y::NTuple{N,Float64})
> quote
> for j in 1:$N, i in eachindex(X)
> X[i][j] = Y[j]
> end
> return X
> end
> end
>
> @generated function testfun2!{N}(X,Y::NTuple{N,Float64})
> quote
> for j in 1:$N, i in eachindex(X)
> X[i][j] = Y[j]
> end
> return X
> end
> end
>
> # Setup for function call
> InnerArrayPts = 3
> OuterArrayPts = 10
> Xinput = [Array{Float64}(InnerArrayPts) for r in 1:OuterArrayPts]
> Yinput = rand(InnerArrayPts)
>
> # Method Error Problem
> testfun1!(Xinput,tuple(Yinput...))
>
> # This works
> testfun2!(Xinput,tuple(Yinput...))
>
> I also tried with the following version of testfun1!() and again got a
> method error.
>
> @generated function
> testfun1!{N}(X::Array{Array{Float64,N},1},Y::NTuple{N,Float64})
> quote
> for j in 1:$N, i in eachindex(X)
> X[i][j] = Y[j]
> end
> return X
> end
> end
>
>
> I am sure i am misunderstanding something quite fundamental and/or missing
> something straightforward...
>
> Thanks,
> Alan
