CODEWITHSUNDEEP
julia
bassen
bassen

Reputation: 525

Julia convention for optional arguments

Say I have a function such as f(x,y) but the y parameter is optional. What is the preferred way to set y as an optional argument? One option that works for me:

function f(x, y=nothing)
    # do stuff 
    if y == nothing
        # do stuff
    else
        # do stuff
    end
    # do stuff
end

But is this the preferred way? I can't set y to a single default value to use in calculation, as there are a number of calculations that are done differently when y is nothing. I could also just have separate functions f(x) and f(x,y) but that seems like too much code duplication.

Upvotes: 18

Views: 13625

Answers (1)

Chris Rackauckas
Chris Rackauckas

Reputation: 19132

This is fine. Note that optional arguments cause dispatch. This means that the if y == nothing (or equivalently, if typeof(y) <: Void), will actually compile away. You'll get two different functions which depend on whether the user gives a value or not. So in the end, the if statement compiles away and it's perfectly efficient to do this.

I will note that the same is not currently true for keyword arguments.

Another way to do this is to have two methods:

f(x)

and

f(x,y)

Whether two methods is nicer than 1 method with an if depends on the problem. Since the if will compile away using type information, there's no difference between these two approaches other than code organization.

Upvotes: 16

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