当包含参数化类型时,在Java中创建泛型类型的实例吗?
这是我的问题的后续行动:
This is a follow-up to my question:
创建的实例当参数化类型通过层次结构传递时,Java中的泛型类型?
为了尝试从包含的类创建新的泛型,我尝试采用Steve B的创建匿名子类的方法:
For attempting to create a new generic from a contained class, I tried to adapt Steve B's approach of creating an anonymous subclass:
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();
}
}
请注意该行 参数化包含= new Parameterized(){};
Please note the line Parameterized contained = new Parameterized(){};
在这里,我正在尝试创建匿名子类,如另一篇文章中的Steve B所建议的那样.但是,我在getTypeParameterClass()方法中收到ClassCastException.这与我的其他帖子中的异常类型相同.那使我认为我可以使用与史蒂夫·B建议的相同的解决方案.
Here I am attempting to create the anonymous subclass, as suggested by Steve B in the other post. However, I get a ClassCastException in the getTypeParameterClass() method. This is the same type of exception as in my other posting. That lead me to think that I could use the same solution as Steve B suggested for that problem.
只要在某类型的类型定义中对E进行了参数化,您要尝试执行的操作就可以起作用.例如:
What you're trying to do can work so long as E is parameterized in a type definition somewhere. For example:
Parameterized<E> pe = new Parameterized<E>();
这将不允许您解析E
,因为它不是类型定义的一部分.另一方面,这是
This will not allow you to resolve E
since it's not part of a type definition. On the otherhand, this:
class StringE extends Parameterized<String> {}
或者这个:
Parameterized<String> ps = new Parameterized<String>(){};
将起作用,因为我们将E的值指定为类型定义的一部分.要解析E
的值,您可以使用 TypeTools :
will work since we're specifying the value of E as part of a type definition. To resolve the value of E
, you might use TypeTools:
Class<?> stringType = TypeResolver.resolveRawArgument(Parameterized.class, ps.getClass());
assert stringType == String.class;