And, as I know, it's not possible to call protocol's implementation in that case
> 16 нояб. 2016 г., в 4:29, Rick Mann via swift-users <swift-users@swift.org>
> написал(а):
>
> Okay. I coudln't find official documentation on this, and I don't currently
> need to do this, but wanted to fully understand it.
>
>> On Nov 15, 2016, at 17:27 , zh ao <owe...@gmail.com> wrote:
>>
>> 'Default' implementation in protocol extension is used as fail safe. You
>> should not consider it like something super class does. If you want it that
>> way, use class inheritance instead.
>>
>> Zhaoxin
>>
>> Get Outlook for iOS
>>
>> _____________________________
>> From: Rick Mann via swift-users <swift-users@swift.org>
>> Sent: 星期三, 十一月 16, 2016 07:51
>> Subject: Re: [swift-users] Attempting to call default protocol
>> implementation crashes Playground
>> To: Dan Loewenherz <d...@lionheartsw.com>
>> Cc: swift-users <swift-users@swift.org>
>>
>>
>> Well, this is a standard protocol default implementation. I was
>> experimenting to see if it was possible to call the default implementation
>> after providing a concrete implementation.
>>
>>> On Nov 15, 2016, at 14:47 , Dan Loewenherz <d...@lionheartsw.com> wrote:
>>>
>>> What are you trying to accomplish here, more concretely?
>>>
>>> My first thought is that you shouldn't implement the same function in both
>>> a protocol extension and a conforming class. Why not just give them
>>> different names and call the function from within the extension instead of
>>> from the class? E.g.
>>>
>>> protocol FooPro {
>>> func _fooFunc()
>>> }
>>>
>>> extension FooPro {
>>> func fooFunc() {
>>> print("fooFunc default")
>>> _fooFunc()
>>> }
>>> }
>>>
>>> class FooClass: FooPro {
>>> func _fooFunc() {
>>> print("fooFunc FooClass")
>>> }
>>> }
>>>
>>> let fc = FooClass()
>>> fc.fooFunc()
>>>
>>> Dan
>>>
>>> On Tue, Nov 15, 2016 at 4:28 PM, Rick Mann via swift-users
>>> <swift-users@swift.org> wrote:
>>> The following gives Xcode 8.1 a very hard time. Eventually I get a Bad
>>> Access on the last line. I'm guessing it's a recursive call. Is there any
>>> way to call the default implementation from a "real" implementation?
>>>
>>> protocol FooPro
>>> {
>>> func fooFunc()
>>> }
>>>
>>> extension FooPro
>>> {
>>> func
>>> fooFunc()
>>> {
>>> print("fooFunc default")
>>> }
>>> }
>>>
>>> class FooClass : FooPro
>>> {
>>> func
>>> fooFunc()
>>> {
>>> (self as FooPro).fooFunc()
>>> print("fooFunc FooClass")
>>> }
>>> }
>>>
>>> let fc: FooPro = FooClass()
>>> fc.fooFunc()
>>>
>>>
>>> Thanks!
>>>
>>>
>>> --
>>> Rick Mann
>>> rm...@latencyzero.com
>>>
>>>
>>> _______________________________________________
>>> swift-users mailing list
>>> swift-users@swift.org
>>> https://lists.swift.org/mailman/listinfo/swift-users
>>>
>>
>>
>> --
>> Rick Mann
>> rm...@latencyzero.com
>>
>>
>> _______________________________________________
>> swift-users mailing list
>> swift-users@swift.org
>> https://lists.swift.org/mailman/listinfo/swift-users
>>
>>
>
>
> --
> Rick Mann
> rm...@latencyzero.com
>
>
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