How about this?
func output(_ separator: String, _ terminator: String, _ items: Any...) {
print(items, separator, terminator)
}
output(",", "\n", "Apple", "Banana")
Zhaoxin
On Mon, Aug 29, 2016 at 3:47 PM, Jin Wang <owenwjowe...@gmail.com> wrote:
> 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 <owe...@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@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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