Question

我有

class Shape implements Comparable <Shape>

和

class Square extends Shape

我写了一个通用方法来查找数组中的最大元素：

public static <S extends Comparable<S>> S findMax(S[] arr)
{
    //blablabla...
    return maxS;
}

这两个电话没有给我任何错误并做了他们应该做的事情：

Shape maxShape = findMax(new Shape[]{new Shape(1), new Shape(2), new Shape(3)});
Square maxSquare = findMax(new Square[]{new Square(1), new Square(2), new Square(3)});

因此，对我而言，由于Shape实现Comparable<Shape>而Square扩展Shape，Square s也应具有可比性，即Square，这似乎是合理的。 1}}以某种方式通过继承自动实现Comparable<Square>（特别是通过继承compareTo(Shape s)）。

然而，根据我的教科书，情况并非如此：这里＆＃34;我们所知道的只是Square implements Comparable<Shape>;因此Square IS-A Comparable<Shape>，但IS-NOT-A Comparable<Square>＆＃34;，它建议使用更好的方法签名：
public static <S extends Comparable<? super S>>。

那为什么我的public static <S extends Comparable<S>>没有给我任何问题？

-----------------------------更新（来源代码）------------- -----------------

public class Shape implements Comparable<Shape>{
    protected int area;
    public Shape (int i)
    {
        this.area=i;
    }

    public String toString()
    {
        return area+"";
    }

    public static void main(String[] args)
    {

        System.out.println("Bigger shape: "+findMax(new Shape[] {new Shape(2),new Shape(3)}));
        System.out.println("Bigger square: "+findMax(new Square[] {new Square(2),new Square(3)}));

    }

    public int getValue()
    {
        return area;
    }

    @Override
    public int compareTo(Shape sh) {
        return Integer.valueOf(area).compareTo(sh.getValue());
    }

    public static <N extends Comparable<N>> N findMax(N[] arr)
    {
        int maxIdx=0;
        for (int i=1; i<arr.length; i++)
            if (arr[i].compareTo(arr[maxIdx])>0)
                maxIdx=i;
        return arr[maxIdx];
    }

}

class Square extends Shape
{
    public Square(int i)
    {
        super(i);
    }

    public int compareTo(Shape sh)
    {
        return Integer.valueOf(area%3).compareTo(sh.getValue()%3);
    }
}

我得到的输出是

Bigger shape: 3
Bigger square: 2

经验教训：原问题的答案是否定的。正如Tagir Valeev所指出的那样，由于findMax的协变性质，允许在Square[] Shape[]上拨打AudioManager.RINGER_MODE_SILENT而无需转让。

Answer 1

实际上你的代码没有编译。 Javac 1.7：

> "C:\Program Files\Java\jdk1.7.0_80\bin\javac.exe" GenericTest.java
GenericTest.java:32: error: method findMax in class GenericTest cannot be applied to given types;
                Square maxSquare = findMax(new Square[]{new Square(1), new Square(2), new Square(3)});
                                   ^
  required: S[]
  found: Square[]
  reason: inferred type does not conform to declared bound(s)
    inferred: Square
    bound(s): Comparable<Square>
  where S is a type-variable:
    S extends Comparable<S> declared in method <S>findMax(S[])
1 error

Javac 1.8：

>"C:\Program Files\Java\jdk1.8.0_40\bin\javac.exe" GenericTest.java
GenericTest.java:32: error: incompatible types: inference variable S has incompatible bounds
                Square maxSquare = findMax(new Square[]{new Square(1), new Square(2), new Square(3)});
                                          ^
    equality constraints: Shape
    upper bounds: Square,Comparable<S>
  where S is a type-variable:
    S extends Comparable<S> declared in method <S>findMax(S[])
1 error

ECJ 3.10.2：

>java -jar org.eclipse.jdt.core_3.10.2.v20150120-1634.jar -source 1.7 GenericTest.java
----------
1. ERROR in C:\projects\Test\src\GenericTest.java (at line 32)
        Square maxSquare = findMax(new Square[]{new Square(1), new Square(2), new Square(3)});
                           ^^^^^^^
Bound mismatch: The generic method findMax(S[]) of type GenericTest is not applicable for the arguments (GenericTest.Square[]). 
The inferred type GenericTest.Square is not a valid substitute for the bounded parameter <S extends Comparable<S>>
----------
1 problem (1 error)

所有编译器都会按预期生成正确的错误消息。如果您将findMax方法声明为public static <S extends Comparable<? super S>> S findMax(S[] arr)，则错误消息会正确消失。

发布完整代码后，

更新问题变得清晰。不同之处在于您没有将findMax的结果分配给变量：

System.out.println("Bigger shape: "+findMax(new Shape[] {new Shape(1),new Shape(3)}));
System.out.println("Bigger square: "+findMax(new Square[] {new Square(3),new Square(2)}));

因此，在这两种情况下，<S>被推断为Shape，Square[]类型为IS-A Shape[]类型。

子类型是否继承＆＃34;使用该特定子类型（Java）参数化的泛型接口？

1 个答案: