嵌套合併運算符的混淆

Question

我最近遇到了這個代碼片段。

let s1: String?? = nil 
(s1 ?? "inner") ?? "outer" // Prints inner 





let st2: String?? = .some(nil) 
(st2 ?? "inner") ?? "outer" // prints outer 

不知道爲什麼（s2 ?? "inner"）返回nil。完全困惑於此。有人能幫我理解原因嗎？

Answer 1

最初不顧結合使用nil合併運算符的：與嵌套可選類型的工作（由於某種原因）時，它可以幫助明確打出來的類型（而不是使用普通?語法糖每個可選「水平」）。例如： -

let s1: Optional<Optional<String>> = nil 
    /* ^^^^^^^^^................^- 'nil' with regard to "outer" optional */ 

let s2: Optional<Optional<String>> = .some(nil) 
    /* ^^^^^^^^^................^- the "outer" optional has value .some(...), i.e, 
            not 'nil' which would be .none. 
       ^^^^^^^^^^^^^^^^-the "inner" optional, on the other hand, has 
            value 'nil' (.none) */ 

使用顯式類型嵌套可選（String??）和分析如上兩種不同的任務，我們就可以着手評估兩者結合nil合併運算上的每個實例調用。顯而易見的是：

let foo1 = (s1 ?? "inner") ?? "outer" // equals "inner" 
     /* ^^- is 'nil', hence the call to 's1 ?? "inner" will coalesce 
       to "inner", which is a concrete 'String' (literal), which 
       means the second nil coalescing operator will not coelesce 
       to "outer" */ 

let foo2 = (s2 ?? "inner") ?? "outer" // equals "outer" 
     /* ^^- is .some(...), hence the call to 's1 ?? "inner" will coalesce 
       to _the concrete value wrapped in 's1'_; namely 'nil', due some, .some(nil). 
       hence, (s1 ?? "inner") results in 'nil', whereafter the 2nd nil 
       coalescing call, 'nil ?? "outer"', will naturally result in 'outer' */ 

對理解略微棘手s2情況下的關鍵是與LHS施加nil合併運算符（左手側）即.some(...)將總是導致在由包裹值即使包裝值本身恰巧是nil（或.none），也是如此。

Optional<SomeType>.some(someThing) ?? anotherThing 
// -> someThing, even if this happens to be 'nil' 

這也很明顯，如果我們選擇看看the stdlib implementation of the nil coalescing operator：

public func ?? <T>(optional: T?, defaultValue: @autoclosure() throws -> T) 
    rethrows -> T { 
    switch optional { 
    case .some(let value): 
    // in your example (s2 and leftmost ?? call), 'T' is Optional<String>, 
    // and 'value' will have the value 'nil' here (which is a valid return for 'T') 
    return value 
    case .none: 
    return try defaultValue() 
    } 
} 

Answer 2

let st2: String?? = .some(nil) 
(st2 ?? "inner") ?? "outer" // prints outer 

不知道爲什麼（S2 ?? 「內部」）返回nil

因爲這是你放在那裏：

let st2: String?? = .some(nil) 
          ^^^ 

比較：

let st2: String?? = .some("howdy") 
(st2 ?? "inner") ?? "outer" // prints howdy