相关疑难解决方法(0)

大家都知道龙延伸Number。那么为什么不编译呢？

以及如何定义方法with，使程序无需任何手工转换就可以编译？

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 …

Why is

public <R, F extends Function<T, R>> Builder<T> withX(F getter, R returnValue) {...}

more strict then

public <R> Builder<T> with(Function<T, R> getter, R returnValue) {...}

This is a follow up on Why is lambda return type not checked at compile time. I found using the method withX() like

.withX(MyInterface::getLength, "I am not a Long")

produces the wanted compile time error:

The type of getLength() from the type BuilderExample.MyInterface is long, this is incompatible with the descriptor's return type: …