[swift-users] Variadic parameter in function type

[Jin Wang]

Hey Zhao,

Thanks for your reply, but then i can’t use the default value `print` which is the print function Swift provides. Any idea?

Cheers,
Jin

> On 29 Aug. 2016, at 5:44 pm, zh ao <owenzx at gmail.com> wrote:
> 
> It is suggested to put ... part at the end.
> 
> Zhaoxin
> 
> Get Outlook for iOS <https://aka.ms/o0ukef>
> 
> 
> 
> On Mon, Aug 29, 2016 at 1:19 PM +0800, "Jin Wang via swift-users" <swift-users at swift.org <mailto:swift-users at swift.org>> wrote:
> 
> Hey guys,
> 
> Can anyone tell me how you handle the following scenario after SE-0111 <https://github.com/apple/swift-evolution/blob/master/proposals/0111-remove-arg-label-type-significance.md> gets implemented?
> 
> let output: (_ items: Any..., _ separator: String, _ terminator: String) -> Void = print
> output("Apple", "Banana", ",", "\n”)
> // Error: Missing argument for parameter #2 in call
> 
> Thanks,
> Jin

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