[英]Swift Type Inference and Protocols with Associated Types
EDIT : I can't understand why in the where clause below - where U.CacheType == T
in AnyCacheable class Swift doesn't treat that statement as a constraint but simply sets T to U.CacheType
. 编辑 :我不明白为什么在下面的where子句中-AnyCacheable类中的
where U.CacheType == T
Swift不会将该语句视为约束,而只是将T设置为U.CacheType
。 Type inference is the worst when things aren't readily apparent :-) 当事情不那么明显时,类型推断是最糟糕的:-)
I am trying to follow Swift's Type erasure discussed here - 我正在尝试遵循此处讨论的Swift的Type擦除-
Specifically the code below: 具体来说,下面的代码:
protocol Cacheable {
associatedtype CacheType
func decode(_ data:Data) ->CacheType?
func encode()->Data?
}
extension String:Cacheable {
func decode(_ data:Data)->String? {
let string = String(data: data, encoding: .utf8)
return string
}
func encode()->Data? {
return data(using: .utf8)
}
}
class AnyCacheable<T>:Cacheable {
private let _encode:()->Data?
private let _decode:(_ data:Data)->T?
init<U:Cacheable>(_ cacheable:U) where U.CacheType == T {
self._encode = cacheable.encode
self._decode = cacheable.decode
}
func decode(_ data:Data)->T? {
return _decode(data)
}
func encode() -> Data? {
return _encode()
}
}
It works perfectly fine if I create a new instance of AnyCacheable
as - 如果我将
AnyCacheable
的新实例创建为-
let cacheable:AnyCacheable = AnyCacheable("Swift")
I don't need to explicitly specify the concrete type of 'T' like let cacheable:AnyCacheable = AnyCacheable<String>("Swift")
我不需要显式指定'T'的具体类型,如
let cacheable:AnyCacheable = AnyCacheable<String>("Swift")
How does Swift infer the concrete type for 'T'? Swift如何推断'T'的具体类型? From the initializer -
从初始化器-
init<U:Cacheable>(_ cacheable:U) where U.CacheType == T {
self._encode = cacheable.encode
self._decode = cacheable.decode
}
I can see that Swift can infer the type for 'U' from the initializer argument (in this case a String type). 我可以看到Swift可以从初始化参数中推断出'U'的类型(在这种情况下为String类型)。 In the where clause 'T' is on rhs.
在where子句中,“ T”在rhs上。 So how does that expression evaluate to true?
那么该表达式如何评估为真?
String is a Cacheable and its decode
returns a String, so its associated type CacheType must be String. String是可缓存的,并且其
decode
返回String,因此其关联类型CacheType必须为String。
AnyCacheable is initialized with a String, which is a Cacheable as required; AnyCacheable使用字符串初始化,该字符串是所需的Cacheable; so its U is String.
所以它的U是String But
U.CacheType
is T. So T is String. 但是
U.CacheType
是T。所以T是String。
To see that this is so, change the definition of String's adoption of Cacheable to this: 要看到是这样,请将String通过Cacheable的采用的定义更改为:
extension String:Cacheable {
func decode(_ data:Data)->Int? {
return 42
}
func encode()->Data? {
return data(using: .utf8)
}
}
Now compile your code and look to see what type you get in the line 现在编译您的代码,看一看得到的类型
let cacheable:AnyCacheable = AnyCacheable("Swift")
It is AnyCacheable<Int>
. 它是
AnyCacheable<Int>
。
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