的Java 8選配的清單收集到列表編輯錯誤

Question

我無法理解的差異

Stream<Optional<Integer>> optionalStream = Stream.of(
       Optional.of(1), 
       Optional.empty(), 
       Optional.of(5)); 

List<Optional<Integer>> optionalList = optionalStream.collect(Collectors.toList()); 

其中一期工程只是罰款之間：

List<Optional<Integer>> optionalList1 = Stream.of(
       Optional.of(1), 
       Optional.empty(), 
       Optional.of(5)).collect(Collectors.toList()); 

當我收到錯誤

Error:(138, 40) java: incompatible types: inference variable T has incompatible bounds 
    equality constraints: java.util.Optional<java.lang.Integer> 
    lower bounds: java.util.Optional<? extends java.lang.Object> 

Answer 1

我已經減少了例子位，並試圖編譯與-XDverboseResolution=all約的類型推斷輸出信息：

final class One { 
    void one() { 
    Stream<Optional<Integer>> optionalStream = Stream.of(Optional.empty()); 
    List<Optional<Integer>> optionalList = optionalStream.collect(Collectors.toList()); 
    } 
} 

final class Two { 
    void two() { 
    List<Optional<Integer>> optionalList1 = 
     Stream.of(Optional.empty()).collect(Collectors.toList()); 
    } 
} 

在Two的情況下，它看起來像的Stream.of遞延實例是連看後續collect之前完成：

... 
Two.java:9: Note: Deferred instantiation of method <T>of(T) 
     Stream.of(Optional.empty()).collect(Collectors.toList()); 
       ^
    instantiated signature: (Optional<Object>)Stream<Optional<Object>> 
    target-type: <none> 
    where T is a type-variable: 
    T extends Object declared in method <T>of(T) 
Two.java:9: Note: resolving method collect in type Stream to candidate 0 
     Stream.of(Optional.empty()).collect(Collectors.toList()); 
... 

（下稱「解決方法collect是collect首提）

沒有target-type限制它;實例化的簽名顯示它是Stream<Optional<Object>>。

如果你看一下對應的輸出One：

... 
One.java:8: Note: Deferred instantiation of method <T>of(T) 
    Stream<Optional<Integer>> optionalStream = Stream.of(Optional.empty()); 
                 ^
    instantiated signature: (Optional<Integer>)Stream<Optional<Integer>> 
    target-type: Stream<Optional<Integer>> 
    where T is a type-variable: 
    T extends Object declared in method <T>of(T) 
... 

它，因爲它知道目標類型得到這一權利。

對於Two，我不能準確地說出爲什麼延遲實例發生在這一點，因爲我對應用類型推斷的方式不夠熟悉。

我認爲這是因爲Stream.of調用不被認爲是多晶硅的表情，但我真的不能相信自己，爲什麼（見編輯歷史有些語無倫次隨筆）。

---

我建議的解決方法是一種提示適用於Optional.empty()，即Optional.<Integer>empty()。這有獲得實際類型的Optional權早些時候推斷的效果，所以它在延遲實例化的時間是已知的，事件雖然目標類型仍是未知數：

final class Three { 
    void three() { 
    List<Optional<Integer>> optionalList1 = 
     Stream.of(Optional.<Integer>empty()).collect(Collectors.toList()); 
    } 
} 

... 
Three.java:9: Note: resolving method of in type Stream to candidate 1 
     Stream.of(Optional.<Integer>empty()).collect(Collectors.toList()); 
      ^
    phase: BASIC 
    with actuals: Optional<Integer> 
    with type-args: no arguments 
    candidates: 
     #0 not applicable method found: <T#1>of(T#1...) 
     (cannot infer type-variable(s) T#1 
      (argument mismatch; Optional<Integer> cannot be converted to T#1[])) 
     #1 applicable method found: <T#2>of(T#2) 
     (partially instantiated to: (Optional<Integer>)Stream<Optional<Integer>>) 
    where T#1,T#2 are type-variables: 
    T#1 extends Object declared in method <T#1>of(T#1...) 
    T#2 extends Object declared in method <T#2>of(T#2) 
Three.java:9: Note: Deferred instantiation of method <T>of(T) 
     Stream.of(Optional.<Integer>empty()).collect(Collectors.toList()); 
       ^
    instantiated signature: (Optional<Integer>)Stream<Optional<Integer>> 
    target-type: <none> 
    where T is a type-variable: 
    T extends Object declared in method <T>of(T) 
... 

Answer 2

Stream.of(...)或Optional.empty()是通用方法。如果你沒有提供類型參數，它會被推斷出來。對於Optional.empty()你會得到Optional<Object>所以Stream.of(Optional.of(1), Optional.empty(), Optional.of(5))將導致Stream<Optional<? extends Object>>。

您可以通過在Optional.<Integer>empty()或Stream.<Optional<Integer>>of(...)中提供類型參數來解決問題。我更喜歡第一個。