我有一个名为Point
的类,它有一个返回Point
数组的neighbors()
方法:
public class Point {
public Point[] neighbors() { /* implementation not shown */ }
}
我有一个Point
的子类,叫做SpecialPoint
,它重写neighbors()
返回一个SpecialPoint
数组而不是Point
数组,我认为这叫做协变返回类型。
public class SpecialPoint extends Point {
public SpecialPoint[] neighbors() { /* implementation not shown */ }
}
在一个单独的类中,我希望通过泛型使用Point
和SpecialPoint
public <P extends Point> P doStuff(P point) {
P[] neighbors = point.neighbors();
// more stuff here including return
}
这不会编译,因为编译器只能保证P
是Point
的某个子类,但不能保证Point
的每个子类都会重写neighbors()
以返回自身的数组,就像我对SpecialPoint
所做的那样,所以Java只知道P#neighbors()
返回Point[]
,而不是P[]
。
我如何保证每个子类都用协变返回类型覆盖neighbors()
,这样我就可以在泛型中使用它?
发布于 2015-04-16 22:03:14
您可以使用一个接口:
public interface Point<P extends Point<P>> {
P[] neighbors();
}
public class SimplePoint implements Point<SimplePoint> {
@Override
public SimplePoint[] neighbors() { /* ... */ }
}
public class SpecialPoint implements Point<SpecialPoint> {
@Override
public SpecialPoint[] neighbors() { /* ... */ }
}
然后:
public <P extends Point<P>> P doStuff(P point) {
P[] neighbors = point.neighbors();
/* ... */
}
如果你仍然需要在实现之间分解代码,那么最好使用抽象类:
public abstract class Point<P extends Point<P>> {
public abstract P[] neighbors();
public void commonMethod() { /* ... */ }
}
public class SimplePoint extends Point<SimplePoint> { /* ... */ }
public class SpecialPoint extends Point<SpecialPoint> { /* ... */ }
发布于 2015-04-16 22:16:00
interface Point
可能会解决您的问题:
public class Test
{
public interface Point {
public Point[] neighbors();
}
public class SpecialPoint implements Point {
public SpecialPoint[] neighbors() { return null; }
}
public class SpecialPoint2 implements Point {
public SpecialPoint2[] neighbors() { return null; }
}
public Point doStuff(SpecialPoint point) {
Point[] neighbors = point.neighbors();
return neighbors[0];
}
public Point doStuff(SpecialPoint2 point) {
Point[] neighbors = point.neighbors();
return neighbors[0];
}
}
https://stackoverflow.com/questions/29676945
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