CODEWITHSUNDEEP
julia
CRquantum
CRquantum

Reputation: 636

How to define a Fortran type equivalent object in Julia

I have a Fortran type I defined as mean_covar like below,

type :: mean_covar
    real(kind=r8), allocatable :: mu(:,:)
    real(kind=r8), allocatable :: sigma(:,:)
    real(kind=r8) :: w
end type

So that I can set some variable as type Mean_covar.

I want to define the same thing in Julia, I know Julia has something called struct, with mutable property its fields can be changed. So I did a very simply thing like,

mutable struct Mean_covar
    mu::Array{Float64,2}
end

Then I define,

musigma = Mean_covar(Array{Float64,2}(undef,2,2))

so I can then set musigma.mu as any 2d array I want.

But I just wonder, does it has to be so cumbersome when defining musigma (need to set the value of each fields)? Can I just simply do

musigma::Mean_covar

But this give me an error,

UndefVarError: musigma not defined

Upvotes: 1

Views: 105

Answers (1)

Przemyslaw Szufel
Przemyslaw Szufel

Reputation: 42244

I am not exactly sure how you want to structure your data but you could do:

struct MeanCovar
    mu::Matrix{Float64}
    MeanCovar(n) = new(zeros(n,n))
end

This could be used as simple as:

julia> m = MeanCovar(2)
MeanCovar([0.0 0.0; 0.0 0.0])

Note that when you know the size of MeanCovar it can be not mutable since you can still mutate the values in mu, as mu hold only reference to the array.

julia> m.mu[1,1] = 66
66

julia> m.mu
2×2 Matrix{Float64}:
 66.0  0.0
  0.0  0.0

Upvotes: 3

Related Questions