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包含参数化类型时在Java中创建泛型类型的实例?

[英]Create instance of generic type in Java when parameterized type is contained?

This is a follow-up to my question:这是我的问题的后续行动:

Create instance of generic type in Java when parameterized type passes through hierarchies? 当参数化类型通过层次结构时,在 Java 中创建泛型类型的实例?

For attempting to create a new generic from a contained class, I tried to adapt Steve B's approach of creating an anonymous subclass:为了尝试从包含的 class 创建一个新的泛型,我尝试采用 Steve B 创建匿名子类的方法:

import java.lang.reflect.Constructor;
import java.lang.reflect.InvocationTargetException;
import java.lang.reflect.ParameterizedType;
import java.lang.reflect.Type;

public class ParameterizedTypeEg {
    ParameterizedTypeEg () {
        ContainsParameterized<String> containString = new ContainsParameterized<String>();
    }
    public class Parameterized<E> {
        Parameterized () {
        }
        public Class<E> getTypeParameterClass() {
            Type type = getClass().getGenericSuperclass();
            ParameterizedType paramType = (ParameterizedType) type;
            return (Class<E>) paramType.getActualTypeArguments()[0];
        }
        public Constructor<E> getTypeParameterConstructor() {
            Constructor<E> constructor = null;
            try {
                constructor = getTypeParameterClass().getConstructor(QueriedColor.class);
            } catch (NoSuchMethodException e) { System.err.println(e); }
            return constructor;
        }           
    }
    class ContainsParameterized<E> {
        ContainsParameterized () {
            Parameterized<E> contained = new Parameterized<E>(){};
            try {
                E element = contained.getTypeParameterConstructor().newInstance();
            }
            catch (InstantiationException e) { System.err.println(e); }
            catch (IllegalAccessException e) { System.err.println(e); }
            catch (InvocationTargetException e) { System.err.println(e); }
        }
    }
    public static void main(String[] args) {
        new ParameterizedTypeEg();
    }
}

Please note the line Parameterized contained = new Parameterized(){};请注意 Parameterized contains = new Parameterized(){};

Here I am attempting to create the anonymous subclass, as suggested by Steve B in the other post.在这里,我正在尝试创建匿名子类,正如 Steve B 在另一篇文章中所建议的那样。 However, I get a ClassCastException in the getTypeParameterClass() method.但是,我在 getTypeParameterClass() 方法中得到了 ClassCastException。 This is the same type of exception as in my other posting.这与我的其他帖子中的异常类型相同。 That lead me to think that I could use the same solution as Steve B suggested for that problem.这使我认为我可以使用与史蒂夫 B 为该问题建议的相同的解决方案。

The anonymous class must have the actual type parameter hard-coded in its declaration for it to work. 匿名类必须在其声明中使用硬编码的实际类型参数才能起作用。

new Parameterized<E>(){};

does not work. 不起作用。 You are just hard-coding "E" (a type variable). 您只是硬编码“ E”(类型变量)。 So when you look at the metadata using reflection all you see is "E" (a type variable). 因此,当您使用反射查看元数据时,您看到的只是“ E”(类型变量)。

You must create a subclass which hard-codes its superclass with an actual class as type argument at compile-time: 您必须创建一个子类,该子类在编译时使用实际的类作为类型参数将其超类硬编码:

new Parameterized<String>(){};

I tried to adapt Steve B's approach of creating an anonymous subclass: 我试图适应史蒂夫·B(Steve B)创建匿名子类的方法:

You missed the whole point of that. 您错过了整个要点。 The point was not to create a subclass. 关键不是要创建子类。 The point was to create a subclass with the actual class of the type argument hard-coded in the superclass . 关键是要创建一个具有超类中硬编码的类型实参类型的实际类的

What you're trying to do can work so long as E is parameterized in a type definition somewhere. 只要在某处的类型定义中对E进行了参数化,您要尝试执行的操作就可以起作用。 For example: 例如:

Parameterized<E> pe = new Parameterized<E>();

This will not allow you to resolve E since it's not part of a type definition. 因为它不是类型定义的一部分,所以这将不允许您解析E On the otherhand, this: 另一方面,这是:

class StringE extends Parameterized<String> {}

or this: 或这个:

Parameterized<String> ps = new Parameterized<String>(){};

will work since we're specifying the value of E as part of a type definition. 将起作用,因为我们将E的值指定为类型定义的一部分。 To resolve the value of E , you might use TypeTools : 要解析E的值,可以使用TypeTools

Class<?> stringType = TypeResolver.resolveRawArgument(Parameterized.class, ps.getClass());
assert stringType == String.class;

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