为什么此类型推断不适用于此Lambda表达式方案？

小编典典

为什么此类型推断不适用于此Lambda表达式方案？

java

我有一个奇怪的场景，在使用lambda表达式时，类型推断无法按预期工作。这是我实际情况的近似值：

static class Value<T> {
}

@FunctionalInterface
interface Bar<T> {
  T apply(Value<T> value); // Change here resolves error
}

static class Foo {
  public static <T> T foo(Bar<T> callback) {
  }
}

void test() {
  Foo.foo(value -> true).booleanValue(); // Compile error here
}

我倒数第二行的编译错误是

未为对象类型定义方法booleanValue（）

如果我将lambda转换为Bar<Boolean>：

Foo.foo((Bar<Boolean>)value -> true).booleanValue();

或者如果我将方法签名更改Bar.apply为使用原始类型：

T apply(Value value);

然后问题就解决了。我希望它能起作用的方式是：

Foo.foo 调用应推断返回类型为 boolean
value在lambda中应该推断为Value<Boolean>。

为什么这种推论不能按预期方式工作？如何更改此API以使其按预期方式工作？

阅读 217

2020-11-13

共1个答案

小编典典

引擎盖下

使用一些隐藏的javac功能，我们可以获得有关正在发生的事情的更多信息：

$ javac -XDverboseResolution=deferred-inference,success,applicable LambdaInference.java 
LambdaInference.java:16: Note: resolving method foo in type Foo to candidate 0
    Foo.foo(value -> true).booleanValue(); // Compile error here
       ^
  phase: BASIC
  with actuals: <none>
  with type-args: no arguments
  candidates:
      #0 applicable method found: <T>foo(Bar<T>)
        (partially instantiated to: (Bar<Object>)Object)
  where T is a type-variable:
    T extends Object declared in method <T>foo(Bar<T>)
LambdaInference.java:16: Note: Deferred instantiation of method <T>foo(Bar<T>)
    Foo.foo(value -> true).booleanValue(); // Compile error here
           ^
  instantiated signature: (Bar<Object>)Object
  target-type: <none>
  where T is a type-variable:
    T extends Object declared in method <T>foo(Bar<T>)
LambdaInference.java:16: error: cannot find symbol
    Foo.foo(value -> true).booleanValue(); // Compile error here
                          ^
  symbol:   method booleanValue()
  location: class Object
1 error

这是很多信息，让我们对其进行分解。

LambdaInference.java:16: Note: resolving method foo in type Foo to candidate 0
    Foo.foo(value -> true).booleanValue(); // Compile error here
       ^
  phase: BASIC
  with actuals: <none>
  with type-args: no arguments
  candidates:
      #0 applicable method found: <T>foo(Bar<T>)
        (partially instantiated to: (Bar<Object>)Object)
  where T is a type-variable:
    T extends Object declared in method <T>foo(Bar<T>)

阶段：方法适用性阶段
实际情况：在
类型参数中传递的实际参数：显式类型参数
候选项：可能适用的方法

实际情况是<none>因为我们的隐式lambda
与适用性无关。

编译器解决您的调用foo来命名的唯一方法foo在Foo。它已经部分实例化到Foo.<Object> foo（因为没有实际值或type-
args），但是可以在推论阶段改变。

LambdaInference.java:16: Note: Deferred instantiation of method <T>foo(Bar<T>)
    Foo.foo(value -> true).booleanValue(); // Compile error here
           ^
  instantiated signature: (Bar<Object>)Object
  target-type: <none>
  where T is a type-variable:
    T extends Object declared in method <T>foo(Bar<T>)

实例化签名：的完全实例化签名foo。这是此步骤的结果（此时，将不再对的签名进行类型推断foo）。
target-type：在其中进行调用的上下文。如果方法调用是分配的一部分，则它将在左侧。如果方法调用本身是方法调用的一部分，则它将是参数类型。

由于您的方法调用是悬空的，因此没有目标类型。由于没有目标类型，因此无法进行进一步的推断，foo并T推断为Object。

分析

编译器在推理期间不使用隐式类型的lambda。在某种程度上，这是有道理的。通常，给定的类型param -> BODY是您将无法编译。如果您确实尝试推断from
的类型，则可能会导致鸡和蛋的类型出现问题。在将来的Java版本中，可能对此进行一些改进。BODY``param``param``BODY

解决方案

Foo.<Boolean> foo(value -> true)

此解决方案提供了一个显式类型参数foo（请注意以下with type- args部分）。这会将方法签名的部分实例化更改为(Bar<Boolean>)Boolean，这就是您想要的。

LambdaInference.java:16: Note: resolving method foo in type Foo to candidate 0
    Foo.<Boolean> foo(value -> true).booleanValue(); // Compile error here
       ^
  phase: BASIC
  with actuals: <none>
  with type-args: Boolean
  candidates:
      #0 applicable method found: <T>foo(Bar<T>)
        (partially instantiated to: (Bar<Boolean>)Boolean)
  where T is a type-variable:
    T extends Object declared in method <T>foo(Bar<T>)
LambdaInference.java:16: Note: resolving method booleanValue in type Boolean to candidate 0
    Foo.<Boolean> foo(value -> true).booleanValue(); // Compile error here
                                    ^
  phase: BASIC
  with actuals: no arguments
  with type-args: no arguments
  candidates:
      #0 applicable method found: booleanValue()

Foo.foo((Value<Boolean> value) -> true)

该解决方案显式地键入您的lambda，从而使其与适用性相关（请注意with actuals以下内容）。这会将方法签名的部分实例化更改为(Bar<Boolean>)Boolean，这就是您想要的。

LambdaInference.java:16: Note: resolving method foo in type Foo to candidate 0
    Foo.foo((Value<Boolean> value) -> true).booleanValue(); // Compile error here
       ^
  phase: BASIC
  with actuals: Bar<Boolean>
  with type-args: no arguments
  candidates:
      #0 applicable method found: <T>foo(Bar<T>)
        (partially instantiated to: (Bar<Boolean>)Boolean)
  where T is a type-variable:
    T extends Object declared in method <T>foo(Bar<T>)
LambdaInference.java:16: Note: Deferred instantiation of method <T>foo(Bar<T>)
    Foo.foo((Value<Boolean> value) -> true).booleanValue(); // Compile error here
           ^
  instantiated signature: (Bar<Boolean>)Boolean
  target-type: <none>
  where T is a type-variable:
    T extends Object declared in method <T>foo(Bar<T>)
LambdaInference.java:16: Note: resolving method booleanValue in type Boolean to candidate 0
    Foo.foo((Value<Boolean> value) -> true).booleanValue(); // Compile error here
                                           ^
  phase: BASIC
  with actuals: no arguments
  with type-args: no arguments
  candidates:
      #0 applicable method found: booleanValue()

Foo.foo((Bar<Boolean>) value -> true)

与上述相同，但口味略有不同。

LambdaInference.java:16: Note: resolving method foo in type Foo to candidate 0
    Foo.foo((Bar<Boolean>) value -> true).booleanValue(); // Compile error here
       ^
  phase: BASIC
  with actuals: Bar<Boolean>
  with type-args: no arguments
  candidates:
      #0 applicable method found: <T>foo(Bar<T>)
        (partially instantiated to: (Bar<Boolean>)Boolean)
  where T is a type-variable:
    T extends Object declared in method <T>foo(Bar<T>)
LambdaInference.java:16: Note: Deferred instantiation of method <T>foo(Bar<T>)
    Foo.foo((Bar<Boolean>) value -> true).booleanValue(); // Compile error here
           ^
  instantiated signature: (Bar<Boolean>)Boolean
  target-type: <none>
  where T is a type-variable:
    T extends Object declared in method <T>foo(Bar<T>)
LambdaInference.java:16: Note: resolving method booleanValue in type Boolean to candidate 0
    Foo.foo((Bar<Boolean>) value -> true).booleanValue(); // Compile error here
                                         ^
  phase: BASIC
  with actuals: no arguments
  with type-args: no arguments
  candidates:
      #0 applicable method found: booleanValue()

Boolean b = Foo.foo(value -> true)

该解决方案为您的方法调用提供了一个明确的目标（请参见target-type下文）。这使deferred-
instantiation可以推断应该使用type参数Boolean代替Object（请参见instantiated signature下文）。

LambdaInference.java:16: Note: resolving method foo in type Foo to candidate 0
    Boolean b = Foo.foo(value -> true);
                   ^
  phase: BASIC
  with actuals: <none>
  with type-args: no arguments
  candidates:
      #0 applicable method found: <T>foo(Bar<T>)
        (partially instantiated to: (Bar<Object>)Object)
  where T is a type-variable:
    T extends Object declared in method <T>foo(Bar<T>)
LambdaInference.java:16: Note: Deferred instantiation of method <T>foo(Bar<T>)
    Boolean b = Foo.foo(value -> true);
                       ^
  instantiated signature: (Bar<Boolean>)Boolean
  target-type: Boolean
  where T is a type-variable:
    T extends Object declared in method <T>foo(Bar<T>)

免责声明

这是正在发生的行为。我不知道这是JLS中指定的内容。我可以四处挖掘，看看是否可以找到指定此行为的确切部分，但是类型推断符号使我头疼。

这也不能完全解释为什么更改Bar为使用原始格式Value可以解决此问题：

LambdaInference.java:16: Note: resolving method foo in type Foo to candidate 0
    Foo.foo(value -> true).booleanValue();
       ^
  phase: BASIC
  with actuals: <none>
  with type-args: no arguments
  candidates:
      #0 applicable method found: <T>foo(Bar<T>)
        (partially instantiated to: (Bar<Object>)Object)
  where T is a type-variable:
    T extends Object declared in method <T>foo(Bar<T>)
LambdaInference.java:16: Note: Deferred instantiation of method <T>foo(Bar<T>)
    Foo.foo(value -> true).booleanValue();
           ^
  instantiated signature: (Bar<Boolean>)Boolean
  target-type: <none>
  where T is a type-variable:
    T extends Object declared in method <T>foo(Bar<T>)
LambdaInference.java:16: Note: resolving method booleanValue in type Boolean to candidate 0
    Foo.foo(value -> true).booleanValue();
                          ^
  phase: BASIC
  with actuals: no arguments
  with type-args: no arguments
  candidates:
      #0 applicable method found: booleanValue()

出于某种原因，将其更改为使用原始Value格式将允许延迟的实例推断T为Boolean。如果我不得不推测的话，我可能会猜想，当编译器尝试将lambda适配到时Bar<T>，它可以通过查看lambda的主体来推断出这T一点Boolean。这意味着我先前的分析是不正确的。编译器
可以对lambda主体执行类型推断，但只能对仅出现在返回类型中的类型变量执行。

2020-11-13