Java泛型Puzzler，扩展一个类并使用通配符 [英] Java Generics Puzzler, extending a class and using wildcards

问题描述

我一直对我的头撞了一阵子，认为也许有些新鲜的眼睛会看到这个问题;感谢您的时间。

  import java.util。*; 
 
 class Tbin< T>扩展ArrayList< T> {} 
 class TbinList< T>扩展ArrayList< Tbin< T>> {} 
 
 class Base {} 
 class Derived extends Base {} 
 $ b public class Test {
 public static void main（String [] args） {
 ArrayList< Tbin<扩展Base>> test = new ArrayList<>（）; 
 test.add（new Tbin< Derived>（））; 
 
 TbinList <？扩展Base> test2 = new TbinList<>（）; 
 test2.add（new Tbin< Derived>（））; 
 
 
 $ / code $ / pre 
 
 
使用Java 8.它在我看来就像直接在 test 中创建容器等同于 test2 中的容器，但编译器说： b.b 
 
  Test.java:15：error：找不到添加的合适方法（Tbin< Derived>）
 test2.add（new Tbin<衍生GT;（））; 
 $ 
  
如何写 Tbin 和 TbinList 所以最后一行可以接受？
请注意，我实际上会添加键入 Tbin  s这就是为什么我在最后一行指定 Tbin< Derived> 的原因。 
 
解决方案
好的，这里是答案： 
 
 
  import java.util。* ; 
 
 class Tbin< T>扩展ArrayList< T> {} 
 class TbinList< T>扩展ArrayList< Tbin<延伸T>> {} 
 
 class Base {} 
 class Derived extends Base {} 
 $ b public class Test {
 public static void main（String [] args） {
 
 TbinList< Base> test3 = new TbinList<>（）; 
 test3.add（new Tbin< Derived>（））; 
 
 
 $ b code 
 $ b $ p $正如我所预料的那样，一旦我看到它。但是很多人为了到这里而th ing不安。 Java的泛型看起来很简单，如果你只看代码。
 
 
 
感谢大家，作为一个声音板。
 
I've been beating my head against this one for awhile and thought that maybe some fresh eyes will see the issue; thanks for your time.
import java.util.*;

class Tbin<T> extends ArrayList<T> {}
class TbinList<T> extends ArrayList<Tbin<T>> {}

class Base {}
class Derived extends Base {}

public class Test {
  public static void main(String[] args) {
    ArrayList<Tbin<? extends Base>> test = new ArrayList<>();
    test.add(new Tbin<Derived>());

    TbinList<? extends Base> test2 = new TbinList<>();
    test2.add(new Tbin<Derived>());
  }
}
Using Java 8. It looks to me like the direct creation of the container in test is equivalent to the container in test2, but the compiler says:
Test.java:15: error: no suitable method found for add(Tbin<Derived>)
    test2.add(new Tbin<Derived>());
         ^
How do I write Tbin and TbinList so the last line is acceptable?

Note that I will actually be adding typed Tbins which is why I specified Tbin<Derived> in the last line.
解决方案
OK, here's the answer:
import java.util.*;

class Tbin<T> extends ArrayList<T> {}
class TbinList<T> extends ArrayList<Tbin<? extends T>> {}

class Base {}
class Derived extends Base {}

public class Test {
  public static void main(String[] args) {

    TbinList<Base> test3 = new TbinList<>();
    test3.add(new Tbin<Derived>());

  }
}
As I expected, kind of obvious once I saw it. But a lot of thrashing around to get here. Java generics seem simple if you only look at working code.

Thanks, everyone, for being a sounding board.

                        
这篇关于Java泛型Puzzler，扩展一个类并使用通配符的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！