获取抽象超类的泛型类型参数的实际类型 [英] Get actual type of generic type argument on abstract superclass
问题描述
我有一个类如下:
public abstract class BaseDao< T extends PersistentObject> {
protected Class< T> getClazz(){
return T.class;
}
// ...
}
但是编译器对类型参数T 。
如何获得 T
?
绝对可以从 Class#getGenericSuperclass()
,因为它没有在运行时定义,但是在编译期间被 FooDao extends BaseDao< ;富>
。
以下是一个启发式示例,您可以在抽象类的构造函数中抽取所需的泛型超类型,并考虑子类的层次结构(以及真实世界的使用在不需要显式提供类型的情况下将其应用于通用 EntityManager
方法): code> public abstract class BaseDao< E extends BaseEntity> {
@PersistenceContext
私人EntityManager em;
私人班级< E>类型;
@SuppressWarnings(unchecked)//用于Class< E>的类型转换。
public BaseDao(){
Type type = getClass()。getGenericSuperclass(); $($)
$ b while(!(typeof instanceof ParameterizedType)||((ParameterizedType)type).getRawType()!= BaseDao.class){
if(type instanceof ParameterizedType){
type =((Class<>)((ParameterizedType)type).getRawType())。getGenericSuperclass();
} else {
type =((Class<>)类型).getGenericSuperclass(); $(b)
$ b this.type =(Class< E>)((ParameterizedType)type).getActualTypeArguments()[0];
}
public E find(Long id){
return em.find(type,id);
}
public List< E> list(){
return em.createQuery(String.format(SELECT e FROM%s e ORDER BY id,type.getSimpleName()),type).getResultList();
}
// ...
}
I have a class like:
public abstract class BaseDao<T extends PersistentObject> {
protected Class<T> getClazz() {
return T.class;
}
// ...
}
But the compiler says to T.class;
: Illegal class literal for the type parameter T
.
How can I get the class of T
?
It's definitely possible to extract it from Class#getGenericSuperclass()
because it's not defined during runtime, but during compiletime by FooDao extends BaseDao<Foo>
.
Here's a kickoff example how you could extract the desired generic super type in the constructor of the abstract class, taking a hierarchy of subclasses into account (along with a real world use case of applying it on generic EntityManager
methods without the need to explicitly supply the type):
public abstract class BaseDao<E extends BaseEntity> {
@PersistenceContext
private EntityManager em;
private Class<E> type;
@SuppressWarnings("unchecked") // For the cast on Class<E>.
public BaseDao() {
Type type = getClass().getGenericSuperclass();
while (!(type instanceof ParameterizedType) || ((ParameterizedType) type).getRawType() != BaseDao.class) {
if (type instanceof ParameterizedType) {
type = ((Class<?>) ((ParameterizedType) type).getRawType()).getGenericSuperclass();
} else {
type = ((Class<?>) type).getGenericSuperclass();
}
}
this.type = (Class<E>) ((ParameterizedType) type).getActualTypeArguments()[0];
}
public E find(Long id) {
return em.find(type, id);
}
public List<E> list() {
return em.createQuery(String.format("SELECT e FROM %s e ORDER BY id", type.getSimpleName()), type).getResultList();
}
// ...
}
这篇关于获取抽象超类的泛型类型参数的实际类型的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!