Hi Neal,
I think rewriting your code to follow a little bit more conventional Julia
style will help you to see where you’ve gone wrong:
type FIR{S, T}
in::Vector{S}
coef::Vector{T}
function FIR(in::Vector{S}, coef::Vector{T})
new(zeros(S, size(in)), coef)
end
end
FIR{S, T}(in::Vector{S}, coef::Vector{T}) = FIR{S, T}(zeros(S, size(in)), coef)
w = zeros(Float64, 10)
f = FIR(w,w)
Note that you must include the outer constructor definition or you will not
have access to the custom inner constructor that you’ve defined. This is one of
the unusual properties of how parametric types work.
Also worth noting that, because of ambiguities, Julia may one day need to
prohibit function invocation with spaces between function names and
parentheses. So it’s better to write FIR(a, b) than to write FIR (a, b).
— John
On Jul 30, 2014, at 8:00 AM, Neal Becker <[email protected]> wrote:
> As a learning exercise, I am trying to code a simple FIR filter.
> As a start, it has 2 fields and a constructor:
>
> type FIR{in_t, coef_t}
> in::Vector{in_t}
> coef::Vector{coef_t}
> FIR (in_::Vector{in_t}, coef_::Vector{coef_t}) = (x=new(); x.in =
> zeros(in_t, size (in_)); x.coef = coef_;)
> end
>
> w = zeros (Float64, 10)
> f = FIR{Float64,Float64}(w,w)
>
> This code executes, but seems very confused:
> julia> typeof(f)
> Array{Float64,1}
>
> Huh? I expected "f" to be of type FIR{Float64,Float64}. Clearly I'm doing
> something very wrong.
>
