julia: type definition contains variable length array

Question

I'd like to define in Julia a composite type which contains a variable length array of another composite type. It's better explained by an example. Let's say I have the type

type p 
    c::Int
    p() = new(0)
end

which I don't really care about. The problem is when I try to define the type

type pp 
   len::Int
   arr::Array{p}(1, len)
end

Obviously I get an error like "len not defined" and I don't know how to fix it. Moreover, how should I define then the constructor of type pp? I'm new to Julia and I'm not even sure if what I'm asking is actually possible.

you can also use FixedSizeArrays.jl to ensure the arr is of length len. — Gnimuc

mbauman mbauman · Accepted Answer · 2016-10-25T22:50:56

The things that go on the right-hand side of the :: need to be types. The expression Array{p}(1, len) isn't a type; it actually creates an array:

julia> len = 3
       Array{p}(1, len)
1×3 Array{p,2}:
 #undef  #undef  #undef

It's uninitialized, but you can see that it's an array itself (and not the type of an array). Its type is simply Array{p,2}. So the minimal fix for your example is simply:

type pp 
   len::Int
   arr::Array{p,2}
end

That's not really what you want, though. You probably just need a vector of p (and not a row matrix, which is what Array{p}(1,len) will create). Also note that Julia's arrays are very full-featured. You don't need to keep track of the length yourself; the arrays do that already.

So I'd probably write a PP type like this:

immutable PP
    arr::Vector{p} # Vector{p} is an alias for Array{p, 1}
end
PP(len::Int) = PP(Vector{p}(len))

julia: type definition contains variable length array

1 Answers