Is it possible to call `fun foo` in <this runnable example> in a way that doesn't provide a value fo...

# getting-started

[Erik]

Is it possible to call fun foo in this runnable example in a way that doesn't provide a value for the first parameter with a default value? The types of the first two parameters happen to be compatible, which is unfortunate (even though they're not the same!)

[Erik]

This is a possible solution

But it's not very nice to have to manually instantiate an array. It beats the purpose of using vararg

[Youssef Shoaib [MOD]]

https://pl.kotl.in/BIUoXKVQ_
Edit: sorry that's the same solution you posted :) I was a minute too late

[Erik]

Thanks, I think it's similar (or the same) as what I did

[Erik]

I guess there's no way around it?

[Youssef Shoaib [MOD]]

I don't think there's a way to make it work even for non-lambda arguments. I'm on mobile so the playground editing experience is horrible, but try the same function where optionalParameter is a Long and the other varargs as Int, it should be a comparable situation

[Erik]

Yeah, probably it is

[dmitriy.novozhilov]

foo(
    optionalParameter = {},
    { println("I want to be the first extra parameter, but I'm the optional parameter") },
    { println("Another one") },
    { println("Yet another one") },
)

[Erik]

Thanks. I guess this is the most concise syntax possible, but still requires a change at the call site

[Erik]

More info: see this thread and this issue.
Let's not continue the discussion here, but there.