抽象继承 (java) 中未经检查的泛型

Question

我收到编译警告："ExampleConsumer.java使用未经检查或不安全的操作return example.distance(other);。如何正确检查类型？显然，我需要强制要求类型相同。

这是我的代码：

示例.java

public abstract class Example<T, U> {
  public T t;
  public U u;
  public Example(T t, U u) {
    this.t = t;
    this.u = u;
  }
  abstract double distance(Example<T, U> other);
}

特殊例子.java

public class SpecialExample extends Example<Integer, Double> {
  public SpecialExample(Integer i, Double d) {
    super(i, d);
  }
  @Override
  double distance(Example<Integer, Double> other) {
    return (double)(t - other.t) + u * other.u;
  }
}

不好的例子.java

public class BadExample extends Example<String, String> {
  public BadExample(String s1, String s2) {
    super(s1, s2);
  }
  @Override
  double distance(Example<String, String> other) {
    return (double)(t.length() + other.t.length()) + (u.length() * other.u.length());
  }
}

示例消费者.java

public class ExampleConsumer<E extends Example> {
  private E example;
  public ExampleConsumer(E example) {
    this.example = example;
  }
  public double combine(E other) {
    return example.distance(other);
  }
}

主.java

class Main {
  public static void main(String[] args) {
    SpecialExample special = new SpecialExample(1, 2.0);
    ExampleConsumer<SpecialExample> consumer = new ExampleConsumer<>(special);
    BadExample bad = new BadExample("foo", "bar");
    consumer.combine(special); // compiles with warning
   // consumer.combine(bad); // doesn't compile = good!
  }
}

Answer 1

这是一个解决方案：

示例消费者.java

public class ExampleConsumer<A, B, E extends Example<A, B>> {
  private E example;
  public ExampleConsumer(E example) {
    this.example = example;
  }
  public double combine(E other) {
    return example.distance(other);
  }
}

主.java

class Main {
  public static void main(String[] args) {
    // ...
    ExampleConsumer<Integer, Double, SpecialExample> consumer = new ExampleConsumer<>(special);
    // ...
  }
}

但我宁愿不必在 Main 中重复双精度/整数类型.java ：/