java - 为什么 Java 不能推断父类(super class)型？

Question

我们都知道龙延伸Number 。那么为什么这不能编译呢？

以及如何定义方法with以便程序无需任何手动转换即可编译？

import java.util.function.Function;

public class Builder<T> {
  static public interface MyInterface {
    Number getNumber();
    Long getLong();
  }

  public <F extends Function<T, R>, R> Builder<T> with(F getter, R returnValue) {
    return null;//TODO
  }

  public static void main(String[] args) {
    // works:
    new Builder<MyInterface>().with(MyInterface::getLong, 4L);
    // works:
    new Builder<MyInterface>().with(MyInterface::getNumber, (Number) 4L);
    // works:
    new Builder<MyInterface>().<Function<MyInterface, Number>, Number> with(MyInterface::getNumber, 4L);
    // works:
    new Builder<MyInterface>().with((Function<MyInterface, Number>) MyInterface::getNumber, 4L);
    // compilation error: Cannot infer ...
    new Builder<MyInterface>().with(MyInterface::getNumber, 4L);
    // compilation error: Cannot infer ...
    new Builder<MyInterface>().with(MyInterface::getNumber, Long.valueOf(4));
    // compiles but also involves typecast (and Casting Number to Long is not even safe):
    new Builder<MyInterface>().with( myInterface->(Long) myInterface.getNumber(), 4L);
    // compiles but also involves manual conversion:
    new Builder<MyInterface>().with(myInterface -> myInterface.getNumber().longValue(), 4L);
    // compiles (compiler you are kidding me?): 
    new Builder<MyInterface>().with(castToFunction(MyInterface::getNumber), 4L);

  }
  static <X, Y> Function<X, Y> castToFunction(Function<X, Y> f) {
    return f;
  }

}

• Cannot infer type argument(s) for <F, R> with(F, R)
• The type of getNumber() from the type Builder.MyInterface is Number, this is incompatible with the descriptor's return type: Long

有关用例，请参阅:Why is lambda return type not checked at compile time

Answer 1

这个表达式:

new Builder<MyInterface>().with(MyInterface::getNumber, 4L);

可以重写为:

new Builder<MyInterface>().with(myInterface -> myInterface.getNumber(), 4L);

考虑方法签名:

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

• R将被推断为 Long
• F将是Function<MyInterface, Long>

并且您传递一个方法引用，该引用将被推断为 Function<MyInterface, Number>这是关键 - 编译器应该如何预测您实际上想要返回 Long来自具有此类签名的函数？ 它不会为您进行向下转型。

自 Number是 Long 的父类(super class)和Number不一定是Long (这就是它无法编译的原因) - 您必须自己显式转换:

new Builder<MyInterface>().with(myInterface -> (Long) myInterface.getNumber(), 4L);

制作F成为Function<MyIinterface, Long>或者像您一样在方法调用期间显式传递通用参数:

new Builder<MyInterface>().<Function<MyInterface, Number>, Number> with(MyInterface::getNumber, 4L);

并且知道R将被视为 Number并且代码将编译。