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-us...@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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