How to add new argument with default value to function which accepts varargs in Kotlin?

Question

Let's say initially I have function:

fun foo(arg1: Int, vararg strings: String) {
}

and there are a lot of calls like this:

foo(1, "a", "b", "c")

Now I want to add additional argument with default value to avoid need changing existing calls:

I tried

fun foo(arg1: Int, arg2: Double = 0.0, vararg strings: String) {
}

But it breaks existing calls:

Type mismatch. Required: Double Found: String

Also I've found this way:

fun foo(arg1: Int, vararg strings: String, arg2: Double = 0.0) {
}

It doesn't break existing calls but if I want to use custom value for arg2 I have to name argument explicitly:

foo(1, "a", "b", "c", arg2 = 0.0)

because following call leads to compile error:

foo(1, "a", "b", "c", 0.0)

Is there better option for my use case ?

"It breaks existing calls." You mean binary incompatible, right? Also what platform are you on? JVM? — Sweeper
– Sweeper, Commented Jan 25, 2024 at 14:39
Oh you're right. I didn't expect this to be source incompatible. — Sweeper
– Sweeper, Commented Jan 25, 2024 at 14:42
Have you tried just refactoring using IntelliJ or some other IDE. Rightclick the function > Refactor > Change Signature. Here you can add new parameters and add default arguments which will be automatically inserted at every call. — Burdui
– Burdui, Commented Jan 25, 2024 at 16:38

Sweeper · Accepted Answer · 2024-01-26 08:12:46Z

2

I would just create a new overload, instead of using parameters with default values.

// add this new overload
fun foo(arg1: Int, arg2: Double, vararg strings: String) {
    // move the implementation to the new overload...
}

fun foo(arg1: Int, vararg strings: String) {
    // the old overload will call the new overload and pass some default values
    foo(arg1, 0.0, *strings)
}

answered Jan 26, 2024 at 8:12

Sweeper

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