因此,我有一个抽象实体Compound和一个抽象服务CompoundService<T extends Compound>

还有另一个服务类,需要访问CompoundService<T extends Compound>的任何实现(任意数字,在编译时未知的实现)。这是1个特定方法所必需的,并且该方法可以使用类参数Class<? extends Compound>

问题是如何将此类安全地映射到相应的服务类?

Map<Class<? extends Compound>, CompoundService<? extends Compound>>之类的东西不起作用,因为我要求CompoundService的通用参数是特定的而不是通配符。

我不确定如何解释这一点。但是compoundService.getById(compoundId)返回类型为CompoundT实例,或者compoundService.save(compound)要求compound类型为T。这意味着

Map<Class<? extends Compound>, CompoundService<? extends Compound>> services = //...;
CompoundService<T> compoundService = services.get(compoundClass);


不起作用,因为不能将CompoundService<? extends Compound>>强制转换为CompoundService<T>

嗯,不确定是否有帮助。问题是类对象Class<? extends Compound>到此“类型”服务类CompoundService<? extends Compound>的映射。

有什么想法可以实现这一目标吗?

最佳答案

我相信这是您想要做的:

abstract class Compound { }

class Compound1 extends Compound { }

abstract class CompoundService<T extends Compound> {
    ...
}

class Compound1Service extends CompoundService<Compound1> {
    ...
}

public class Test {

    static Map<Class<? extends Compound>, CompoundService<? extends Compound>> serviceMap = new HashMap<Class<? extends Compound>, CompoundService<? extends Compound>>();

    public static <T extends Compound> void main(String[] args) {
        serviceMap.put(Compound1.class, new Compound1Service());
        CompoundService<Compound1> service = getServiceFromMap(Compound1.class);
        System.out.println(service.getClass());
    }

    public static <T extends Compound> CompoundService<T> getServiceFromMap(Class<T> clazz) {
        return(CompoundService<T>)serviceMap.get(clazz);
    }

}


我认为您不会比这更好,并且可以避免由于类型擦除而导致的类型转换

07-26 07:00