错误的通用返回类型

Question

让我举个简单的例子:

public class Main {

    public static abstract class Food { 
    }

    public static abstract class Fruit extends Food {
        public String getJuice(){
            return "juice";
        }
    }

    public static abstract class Container<T extends Food> {
        private T food;
        public final T get(){
            return food;
        }
    }

    public static abstract class Crate<T extends Fruit> extends Container<T> {
    }

    public static void Bar(Crate crate) {
        Food  f  = crate.get(); //compiles fine
      //Fruit f2 = crate.get(); //can't compile
    }                           
}                               

当给出原始类型时,crate.get()返回Food而不是Fruit

• 注意:Crate被声明为Crate<T extends Fruit> extends Container<T>使Crate<Food>被禁止的.

我只是很好奇:为什么方法T get()不返回Fruit？为什么Crate<Fruit>需要？

Answer 1

发生这种情况是因为get声明了Container.当使用原始类型Crate(具有对其所有(继承)方法应用擦除的效果)时,签名public T get()被删除public Food get(),因为它的声明类T具有上限.Food

虽然Crate缩小了上限T,恰好是返回类型get,但它不会覆盖该方法.如果是这种情况,您会看到不同的行为:

public static abstract class Container<T extends Food> {
    private T food;
    public T get() {
        return food;
    }
}

public static abstract class Crate<T extends Fruit> extends Container<T> {

    @Override
    public T get() {
        return super.get();
    }
}

public static void bar(Crate crate) {
    Food  f  = crate.get(); // compiles fine
    Fruit f2 = crate.get(); // also compiles
}

现在,public T get()被删除了public Fruit get(),因为它已被重新声明在T具有上限的类中Fruit.