java - 获取泛型类的 .class 对象

Question

我想保存泛型值的类型，因为我无法在运行时获取它:

public class A<T> {
    private final Class<T> genericType;
    public A(Class<T> genericType) {
        this.genericType = genericType;
    }

    public Class getGenericType() {
        return genericType;
    }
}

现在要创建子类，我按如下方式使用它:

public class B extends A<String> {
    public B() {
        super(String.class);
    }
}

注意 super() 的参数类型匹配(通过编译时间检查)到 A 的泛型类型。
这很好用。但是如果我想用 Map 来拥有它，我就无法获得正确的类对象:

public class C extends A<Map<String, String>> {
    public C() {
        super(Map.class); // does not match Map<String,String>
        super(Map<String,String>.class) // no valid java expression, i dont know what 
    }
}

Sooo有人有帮助我摆脱这种痛苦的提示吗？
我目前能做的最好的事情就是放弃 A 中的强类型:

public class A<T> {
    // old: private final Class<T> genericType;
    private final Class genericType;  // note the missing generic
    public A(Class genericType) {     // here as well
        this.genericType = genericType;
    }

    public Class getGenericType() {
        return genericType;
    }
}

Answer 1

我不确定这是否满足您的要求，但您可以执行以下类似操作，请参阅 How to get the class of a field of type T?

import java.lang.reflect.*;
import java.util.*;


public class GenericTypeTest{

     public static void main(String []args){
        B b = new B();
        System.out.println("B is a " + b.getGenericType());
        C c = new C();
        System.out.println("C is a " + c.getGenericType());
     }
}


class A<T> {
      public Class getGenericType() {
          Object genericType =  ((ParameterizedType) getClass().getGenericSuperclass()).getActualTypeArguments()[0];
           if(genericType instanceof ParameterizedType){
               genericType = ((ParameterizedType)genericType).getRawType();
           }
        return (Class<T>) genericType;
    }
}



class B extends A<String> {
}

class C extends A<Map<String,String>> {
}

这将得到类似的输出

B is a class java.lang.String
C is a interface java.util.Map