Java中的泛型：如何定义具有多个不同类型的函数

Question

@Service
public class Animal {
   public String name;
}
@Service
public class Dog extends Animal {
   public String name;
}

@Service
public class Cat extends Animal {
   public String name;
}

In spring boot project, I wanna obtain one specific bean by using ApplicationContext provided by spring framework, here is a simple example i write to illustrate:

@Component
public class AnimalLocator implements ApplicationContextAware {
   private static ApplicationContext applicationContext;
   @Override
   public void setApplicationContext(ApplicationContext applicationContext) throws BeansException {
      if (PayServiceLocator.applicationContext == null) {
         PayServiceLocator.applicationContext = applicationContext;
      }
   }

   public <T extends Animal> T getService(String name) {
      if("Cat".equals(name) {
        applicationContext.getBean(name, Cat.class);
      }
      if("Dog".equals(name) {
        applicationContext.getBean(name, Dog.class);
      }
   }
}

However, exceptions was prompted by the compiler:

the mosaic part should be Dog or cat. I thought it should work, since T has already extended the Animal class, but it doesn't ,so does anyone has any ideas about it? Thank you!

Answer 1

Since you're using bean class to access bean instance, it's straight forward to pass class as parameter:

public <T extends Animal> T getPayService(String name, Class<T> payClass) {
   return applicationContext.getBean(name, payClass);
}

Answer 2

T in getPayService will extend Animal , of course. This means that code calling it with another type won't compile:

Fruit fruit = animalLocator.getPayService("Banana")

To illustrate your current problem , look at this:

Cat cat = animalLocator.getPayService("Dog");

T is Cat in this case, but your code would be returning a Dog .

To circumvent the compiler error, you can add a type cast:

return (T) applicationContext.getBean(...

But this would still not be safe because the compiler will still be unable to guarantee that the actual return type will be what T is in the context of the caller at runtime, and the caller would be having a class cast exception.

If we can assume that getBean is a safe call, then you should change your method to this implementation:

public <T extends Animal> T getPayService(String name, Class<T> cls) {
   return applicationContext.getBean(name, cls);
}

This doesn't change a lot from the caller's perspective, but hinges on the fact (or assumption) that applicationContext.getBean(name, cls); will return an object of type T . This means your code is as type-safe as getBean is, but the compiler is happy with that.

Answer 3

You can Autowired all your Animal instances in a Map instead of coding your if/else :

@Service("Animal")
public class Animal {
    public String name;
}
@Service("Dog")
    public class Dog extends Animal {
}
@Service("Cat")
    public class Cat extends Animal {
}

And in your AnimalLocator :

@Component
public class AnimalLocator {

   @Autowired
   private Map<String,Animal> animals;

   public <T extends Animal> T getService(String name) {
      return this.animals.get(name);
   }
}