Getting a reference to an instance variable with a default parameter in Swift

Question

So given the following code, how does one get a reference to a function that takes a parameter with a default value and invoke the reference with the default value?

class Test {
  func doIt() { print("Done") }
  func doIt(_ adjective: String = "better") {
    print("Done \(adjective)")
  }
}


let t = Test()
let fn1 = t.doIt as () -> Void
let fn2 = t.doIt as (String) -> Void

fn1() // Works
fn2() // Does not work; requires parameter

I also tried the following

let fn2 = t.doIt as (String?) -> Void

But that also does not work. Any ideas? I'd like to invoke fn2() and get the printed result "Done better"

`fn2` is a closure, and closures cannot have default values for their parameters. — Martin R, Jul 14 '18 at 07:01
Possible duplicate of [Why Swift throws error when using optional param in closure func?](https://stackoverflow.com/q/45258621/1187415) — Martin R, Jul 14 '18 at 07:10

score 0 · Answer 1 · edited Jul 14 '18 at 09:41

0

I think you already got what you wanted, a reference to that function.

Since the type is (String) -> Void, it expects String as an input. So you have to pass it to call.

edited Jul 14 '18 at 09:41

pacification

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answered Jul 14 '18 at 06:54

Bista

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Getting a reference to an instance variable with a default parameter in Swift

1 Answers1