为什么编译时不检查 lambda 返回类型？

Question

使用的方法引用具有返回类型Integer。String但在下面的示例中，不兼容是允许的。如何修复方法with声明以获得方法引用类型安全而无需手动转换？import java.util.function.Function;public class MinimalExample {  static public class Builder<T> {    final Class<T> clazz;    Builder(Class<T> clazz) {      this.clazz = clazz;    }    static <T> Builder<T> of(Class<T> clazz) {      return new Builder<T>(clazz);    }    <R> Builder<T> with(Function<T, R> getter, R returnValue) {      return null; //TODO    }  }  static public interface MyInterface {    Integer getLength();  }  public static void main(String[] args) {// missing compiletimecheck is inaceptable:    Builder.of(MyInterface.class).with(MyInterface::getLength, "I am NOT an Integer");// compile time error OK:     Builder.of(MyInterface.class).with((Function<MyInterface, Integer> )MyInterface::getLength, "I am NOT an Integer");// The method with(Function<MinimalExample.MyInterface,R>, R) in the type MinimalExample.Builder<MinimalExample.MyInterface> is not applicable for the arguments (Function<MinimalExample.MyInterface,Integer>, String)  }}

Answer 1

在第一个示例中，和MyInterface::getLength分别帮助"I am NOT an Integer"解析通用参数T和。RMyInterfaceSerializable & Comparable<? extends Serializable & Comparable<?>>

// it compiles since String is a Serializable

Function<MyInterface, Serializable> function = MyInterface::getLength;

Builder.of(MyInterface.class).with(function, "I am NOT an Integer");

MyInterface::getLength并不总是 a ，Function<MyInterface, Integer>除非您明确这么说，这会导致编译时错误，如第二个示例所示。

// it doesn't compile since String isn't an Integer

Function<MyInterface, Integer> function = MyInterface::getLength;

Builder.of(MyInterface.class).with(function, "I am NOT an Integer");

Answer 2

类型推断在这里发挥了作用。R考虑方法签名中的泛型：

<R> Builder<T> with(Function<T, R> getter, R returnValue)

在所列情况中：

Builder.of(MyInterface.class).with(MyInterface::getLength, "I am NOT an Integer");

的类型R被成功推断为

Serializable, Comparable<? extends Serializable & Comparable<?>>

并且 aString确实通过这种类型暗示，因此编译成功。

---

要显式指定 的类型R并找出不兼容性，只需将代码行更改为：

Builder.of(MyInterface.class).<Integer>with(MyInterface::getLength, "not valid");

Answer 3

这是因为你的泛型类型参数R可以被推断为 Object，即以下编译：

Builder.of(MyInterface.class).with((Function<MyInterface, Object>) MyInterface::getLength, "I am NOT an Integer");

Answer 4

这个答案基于其他答案，这些答案解释了为什么它不能按预期工作。

解决方案

下面的代码通过将双函数“with”拆分为两个连贯函数“with”和“returning”来解决该问题：

class Builder<T> {

...

class BuilderMethod<R> {

  final Function<T, R> getter;

  BuilderMethod(Function<T, R> getter) {

    this.getter = getter;

  }

  Builder<T> returning(R returnValue) {

    return Builder.this.with(getter, returnValue);

  }

}

<R> BuilderMethod<R> with(Function<T, R> getter) {

  return new BuilderMethod<>(getter);

}

...

}

MyInterface z = Builder.of(MyInterface.class).with(MyInterface::getLength).returning(1L).with(MyInterface::getNullLength).returning(null).build();

System.out.println("length:" + z.getLength());

// YIPPIE COMPILATION ERRROR:

// The method returning(Long) in the type BuilderExample.Builder<BuilderExample.MyInterface>.BuilderMethod<Long> is not applicable for the arguments (String)

MyInterface zz = Builder.of(MyInterface.class).with(MyInterface::getLength).returning("NOT A NUMBER").build();

System.out.println("length:" + zz.getLength());

（有点陌生）