为什么此类型推断不适用于此Lambda表达式方案？

问题描述：

我有一个奇怪的场景，类型推断不能正常工作，因为我在使用lambda表达式时会发现。这是我的真实场景的近似值：

I have a weird scenario where type inference isn't working as I'd expect when using a lambda expression. Here's an approximation of my real scenario:

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
}

我在第二行到最后一行的编译错误是

The compile error I get on the second to last line is

方法booleanValue（）未定义类型Object

The method booleanValue() is undefined for the type Object

如果我将lambda转换为 Bar< Boolean> ：

if I cast the lambda to Bar<Boolean>:

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

或者如果我更改 Bar.apply $ c的方法签名$ c>使用原始类型：

or if I change the method signature of Bar.apply to use raw types:

T apply(Value value);

然后问题就消失了。我期望它的工作方式是：

then the problem goes away. The way I'd expect this to work is that:

Foo.foo call应在lambda中推断返回类型 boolean

value 推断为值< Boolean> 。

Foo.foo call should infer a return type of boolean
value in the lambda should be inferred to Value<Boolean>.

为什么此推理不能按预期工作？如何更改此API以使其按预期工作？

Why doesn't this inference work as expected and how can I change this API to make it work as expected?

答

在引擎盖下

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：

Under the Hood

Using some hidden javac features, we can get more information about what's happening:

$ 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，让我们把它分解。

This is a lot of information, let's break it down.

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

actuals：传递的实际参数

type-args：显式类型参数

候选人：适用的方法

实际值是< none> 因为我们隐式输入的lambda不是与适用性相关。

actuals is <none> because our implicitly typed lambda is not pertinent to applicability.

编译器解析你的将 foo 调用到 Foo 中名为 foo 的唯一方法。它已被部分实例化为 Foo。< Object> foo （因为没有实际值或类型参数），但是可以在延迟推理阶段改变。

The compiler resolves your invocation of foo to the only method named foo in Foo. It has been partially instantiated to Foo.<Object> foo (since there were no actuals or type-args), but that can change at the deferred-inference stage.

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： foo 的完全实例化签名。这是此步骤的结果（此时不再对 foo 的签名进行类型推断。）
target-type：the正在进行调用的上下文。如果方法调用是赋值的一部分，则它将是左侧。如果方法调用本身是方法调用的一部分，那么它将是参数类型。

instantiated signature: the fully instantiated signature of foo. It is the result of this step (at this point no more type inference will be made on the signature of foo).
target-type: the context the call is being made in. If the method invocation is a part of an assignment, it will be the left hand side. If the method invocation is itself part of a method invocation, it will be the parameter type.

由于方法调用是悬空的，因此没有目标类型。由于没有目标类型，因此无法在 foo 上进行更多推断，并且 T 被推断为对象。

Since your method invocation is dangling, there is no target-type. Since there is no target-type, no more inference can be done on foo and T is inferred to be Object.

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 param - > BODY ，在 param 的类型之前，您将无法编译 BODY 。如果您尝试从 BODY 推断 param 的类型，则可能会导致鸡蛋类型问题。在未来的Java版本中，可能会对此进行一些改进。

The compiler does not use implicitly typed lambdas during inference. To a certain extent, this makes sense. In general, given param -> BODY, you will not be able to compile BODY until you have a type for param. If you did try to infer the type for param from BODY, it might lead to a chicken-and-egg type problem. It's possible that some improvements will be made on this in future releases of Java.

Foo。< Boolean> foo（value - > true）

此解决方案为 foo $ c $提供显式类型参数c>（注意下面的带有type-args 部分）。这会将方法签名的部分实例化更改为（Bar< Boolean>）布尔值，这就是您想要的。

This solution provides an explicit type argument to foo (note the with type-args section below). This changes the partial instantiation of the method signature to (Bar<Boolean>)Boolean, which is what you want.

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）

此解决方案显式键入您的lambda，允许它与适用性相关（注意下面有实际值）。这会将方法签名的部分实例化更改为（Bar< Boolean>）布尔值，这就是您想要的。

This solution explicitly types your lambda, which allows it to be pertinent to applicability (note with actuals below). This changes the partial instantiation of the method signature to (Bar<Boolean>)Boolean, which is what you want.

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>）值 - > true）

与上述相同，但味道略有不同。

Same as above, but with a slightly different flavor.

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）

此解决方案为您的方法调用提供了一个明确的目标（参见目标类型如下）。这允许延迟实例化推断类型参数应该是布尔值而不是对象（参见实例化签名如下）。

This solution provides an explicit target for your method call (see target-type below). This allows the deferred-instantiation to infer that the type parameter should be Boolean instead of Object (see instantiated signature below).

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。我不知道这是否是JLS中指定的内容。我可以四处寻找，看看我是否能找到指定此行为的确切部分，但类型推断表示法令我头疼。

这也无法完全解释为什么要更改 Bar 使用原始值将解决此问题：

This also doesn't fully explain why changing Bar to use a raw Value would fix this issue:

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值允许延迟实例化推断 T 是布尔值。如果我不得不推测，我猜想当编译器试图将lambda拟合到 Bar< T> 时，它可以推断出 T 是 Boolean 。这意味着我之前的分析是不正确的。编译器可以在lambda的主体上执行类型推断，但仅限于仅在返回类型中出现的类型变量。

For some reason, changing it to use a raw Value allows the deferred instantiation to infer that T is Boolean. If I had to speculate, I would guess that when the compiler tries to fit the lambda to the Bar<T>, it can infer that T is Boolean by looking at the body of the lambda. This implies that my earlier analysis is incorrect. The compiler can perform type inference on the body of a lambda, but only on type variables that only appear in the return type.

为什么此类型推断不适用于此Lambda表达式方案？

在引擎盖下

Under the Hood

免责声明

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