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Java泛型-带有可选包装泛型的意外情况

[英]Java generics - unexpected situation with Optional wrapping generics

I have faced following situation: 我已经遇到以下情况:

import com.google.common.base.Optional;

public class CovarianceTest {

    class Race {
    }

    class Dog<Race> {
    }

    public Optional<Dog<? extends Race>> getDog() {
        Dog<? extends Race> dogWithSomeRace = new Dog();
        return Optional.of(dogWithSomeRace);
    }
}

Situation like this causes compilation problem. 这样的情况会导致编译问题。 The IDE says: IDE说:

Incompatible types.
Required: Optional<Dog<? extends Race>>
Found: Optional<? extends Dog<? extends Race>

Which is clearly wrong. 这显然是错误的。

Could someone explain why this is happening? 有人可以解释为什么会这样吗?

It seems you confused the class' generic type declaration with the boundary declaration of a generic type argument. 似乎您将类的泛型类型声明与泛型类型参数的边界声明混淆了。

In your case: 在您的情况下:

class Race {}

class Dog<Race> {} // "Race" is declared as a generic type in dog. It's not referring to your Race class in any way.

Your definition of Dog is equivalent to 您对Dog定义等同于

class Dog<E> {}

If you want to set a subclass of Race to be a generic boundary for Dog , then do for example 如果要将Race的子类设置为Dog的通用边界,请执行例如

class Dog<E extends Race> {}

You could also refine the method's generic declarations: 您还可以优化方法的泛型声明:

public Optional<Dog<? extends Race>> // this could be shortend to Dog<?>, because Dog's boundary is already limited to Race in the class signature
                                     getDog() {
    Dog<? extends Race> dogWithSomeRace // This can be shortend aswell to Dog<?> 
                                        = new Dog(); // Missing type declaration in the constructor call 
    return Optional.of(dogWithSomeRace);
}

This is your code: 这是您的代码:

  1. Compiler thinks Race (in class Dog<Race> ) is a type parameter that you will use for further parametrization of class Dog . 编译器认为Race (在Dog<Race>类中)是一个类型参数,将用于进一步对class Dog进行参数化。
  2. In the assignment Dog<? extends Race> dogWithSomeRace = new Dog(); 在作业Dog<? extends Race> dogWithSomeRace = new Dog(); Dog<? extends Race> dogWithSomeRace = new Dog(); this is considered as not best practice. 这被认为不是最佳实践。

在此处输入图片说明

I believe the code below is what you are trying to do. 我相信下面的代码就是您要尝试执行的操作。 Use T as type parameter. 使用T作为类型参数。 Here we go: 开始了:

import com.google.common.base.Optional;

public class CovarianceTest {

    class Race {
    }

    class Dog<T> {
        void whatever(T t) {
            System.out.println(t);
        }
    }

    public Optional<Dog<? extends Race>> getDog() {
        Dog<? extends Race> dogWithSomeRace = new Dog<Race>();
        return Optional.of(dogWithSomeRace);
    }

    public static void main(String[] args) {
        System.out.println((new CovarianceTest()).getDog().isPresent());
        System.out.println((new CovarianceTest()).getDog().get().getClass());
    }
}

This prints nicely: 打印效果很好:

true 真正
class CovarianceTest$Dog 类CovarianceTest $ Dog

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