In Julia: initialize fields with nothing

Question

I have a mutable struct with optional fields like this:

mutable struct MyStruct
    field1::Union{Int, Nothing}
    field2::Union{String, Nothing}
    field3::Union{Int, Nothing}
    field4::Union{String, Nothing}
    # ...
end

I can now write a default constructor which initializes the fields with nothing:

MyStruct() = MyStruct(nothing, nothing, nothing, nothing)

This is not so nice when my struct has many fields. Also, I have to count the fields to get the constructor call with all the 'nothings' correct in this case. Is there a better way to do that?

Depending on the field content, I want to call different functions later:

if mystruct.field1 == nothing
  do_this()
else
  do_that()
end

I'm interested in the use case, but I'd also like to point out that last example is better suited to using parametric functions, ie f(::Nothing) = ...#something ; f(_) = ... #other thing — stillearningsomething

hckr hckr · Accepted Answer · 2018-10-23T15:27:01

You can use fieldcount function to achieve that. This function gives you the number of fields of that an instance of given type would have. Here is an example containing a mutable struct and an outer constructor.

julia> mutable struct Foo
           x
           y
           z
       end

julia> Foo() = Foo(ntuple(x->nothing, fieldcount(Foo))...); # you can also fill an array and use `...`

julia> Foo()
Foo(nothing, nothing, nothing)

In Julia: initialize fields with nothing

1 Answers