请问一下，在实现 spring 的 FactoryBean 时，如何得到类成员方法里某个局部变量的真实类型？

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rt,代码逻辑大概是这样：

public class MyFactoryBean implements FactoryBean<MyBean> { private String myProperty; public void setMyProperty(String myProperty) { this.myProperty = myProperty; } @Override public MyBean getObject() throws Exception { return createBean(); } public MyBean createBean(){ MyBean myBean = new MyBean(); myBean.setProperty(myProperty); //下面 UserListener 调试时，UserListener 对象的类型是“MyFactoryBean$$Lambda$400/0x0000000800e9b558@4225” UserListener<User1> userListener=new UserListener<User1>(){ @Override public void active(){ //doSomething } }; Cache.addUserListener(userListener); return myBean; } @Override public Class<?> getObjectType() { return MyBean.class; } @Override public boolean isSingleton() { return true; } }

在传递UserListener时，发现传递的不是 UserListener 实例的具体类型(UserListener 是一个泛型接口)，而是一个

MyFactoryBean$$Lambda$400/0x0000000800e9b558@4225

代理对象,请问有什么方式能直接得到该类具体的类型吗？除了把对象类型一并传过去

 Cache.addUserListener(userListener,User1.class);

ps: 哈哈，框架还不是很熟，大家尽管吐槽指正，谢谢大家回复！

Supplement 1 May 23, 2023

哈哈，问了下chatgpt，它说runtime时没法直接得到这个泛型，提供的思路跟我说的基本一样，就是在添加时，把类型传过去。

I understand. In Java, due to type erasure, it is not possible to directly obtain the type parameter T at runtime. However, you can work around this limitation by passing a Class<T> object as a parameter to the addUserListener method. Here's an example of how you can modify the Cache class to achieve this: public class Cache { public static <T extends User> void addUserListener(UserListener<T> userListener, Class<T> userType) { // implementation of the method goes here System.out.println("User type is: " + userType.getSimpleName()); } }

4 replies 2023-05-23 01:25:17 +08:00

letsky

May 21, 2023

试试 Spring 的这个类，https://docs.spring.io/spring-framework/docs/current/javadoc-api/org/springframework/core/ResolvableType.html

```java
private HashMap<Integer, List<String>> myMap;

public void example() {
ResolvableType t = ResolvableType.forField(getClass().getDeclaredField("myMap"));
t.getSuperType(); // AbstractMap<Integer, List<String>>
t.asMap(); // Map<Integer, List<String>>
t.getGeneric(0).resolve(); // Integer
t.getGeneric(1).resolve(); // List
t.getGeneric(1); // List<String>
t.resolveGeneric(1, 0); // String
}
```

luckykev1n

May 21, 2023

不考虑 spring 的话，可以用反射
public class Main {
public static void main(String[] args) {
UserListener<String> userListener = new UserListener<String>() {};
// 获取实际类型
Class<? extends UserListener> listenerClass = userListener.getClass();

// 获取实际范型
Type[] interfaces = listenerClass.getGenericInterfaces();
Type listenerType = interfaces[0];
ParameterizedType paraType = (ParameterizedType) listenerType;
Type[] actualTypes = paraType.getActualTypeArguments();
Class<?> actualType = (Class<?>) actualTypes[0];

// 打印实际范型
System.out.println(actualType);

}
}

luckykev1n

May 21, 2023

1 楼的这个 ResolvableType ，其实也是 Spring 基于 java 的反射实现的，
```
public ResolvableType[] getGenerics() {
if (this == NONE) {
return EMPTY_TYPES_ARRAY;
}
if (this.generics == null) {
if (this.type instanceof Class) {
Class<?> typeClass = (Class<?>) this.type;
this.generics = forTypes(SerializableTypeWrapper.forTypeParameters(typeClass), this.variableResolver);
}
else if (this.type instanceof ParameterizedType) {
Type[] actualTypeArguments = ((ParameterizedType) this.type).getActualTypeArguments();
ResolvableType[] generics = new ResolvableType[actualTypeArguments.length];
for (int i = 0; i < actualTypeArguments.length; i++) {
generics[i] = forType(actualTypeArguments[i], this.variableResolver);
}
this.generics = generics;
}
else {
this.generics = resolveType().getGenerics();
}
}
return this.generics;
}

OldCarMan

May 23, 2023

@letsky 谢谢回复
@luckykev1n 谢谢回复，你 2 楼回复的：
ParameterizedType paraType = (ParameterizedType) listenerType;
类型无法转换（ java.lang.ClassCastException: class java.lang.Class cannot be cast to class java.lang.reflect.ParaeterizedType ）。

1.

可能我没描述好，我说的是，我在另外一个类 Cache.java 里面接收 UserListener 时，无法获取到泛型接口 UserListener<T extends User>相应的泛型 T 类型:
public class Cache{
-----

public static void addUserListener(UserListener<T extends User> userListener){

在这里无法直接获取得到 userListener 的相应的泛型 T 类型

}

-----

}