Julia convention for optional arguments

Question

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.

Chris Rackauckas Chris Rackauckas · Accepted Answer · 2017-02-28T03:52:32

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.

Julia convention for optional arguments

1 Answers