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 SerializableFunction<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 IntegerFunction<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>&nbsp;Builder<T>&nbsp;with(Function<T,&nbsp;R>&nbsp;getter,&nbsp;R&nbsp;returnValue)在所列情况中：Builder.of(MyInterface.class).with(MyInterface::getLength,&nbsp;"I&nbsp;am&nbsp;NOT&nbsp;an&nbsp;Integer");的类型R被成功推断为Serializable,&nbsp;Comparable<?&nbsp;extends&nbsp;Serializable&nbsp;&&nbsp;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,&nbsp;"not&nbsp;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,&nbsp;Object>)&nbsp;MyInterface::getLength,&nbsp;"I&nbsp;am&nbsp;NOT&nbsp;an&nbsp;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> {&nbsp; final Function<T, R> getter;&nbsp; BuilderMethod(Function<T, R> getter) {&nbsp; &nbsp; this.getter = getter;&nbsp; }&nbsp; Builder<T> returning(R returnValue) {&nbsp; &nbsp; return Builder.this.with(getter, returnValue);&nbsp; }}<R> BuilderMethod<R> with(Function<T, R> getter) {&nbsp; 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）