如果我正在为通用类编写包装器,则我尝试将其嵌入到构造中(如this SO question所示):
class Foo<T> {
private T value;
private Class<T> type;
public Foo(Class<T> type, T value) {
this.value = value;
this.type = type;
}
public T getValue() {
return value;
}
public Class<T> getType() {
return type;
}
}
我有一个
Foo
实例的列表,我想将它们转换成FooWrapper
的列表,方法如下:List<Foo<?>> someListOfFoos = ...
List<FooWrapper<?>> fooWrappers = someListOfFoos
.stream()
.map(foo -> FooWrapper.from(foo))
.collect(Collectors.toList());
构建每个
someListOfFoos
时,有什么方法可以恢复FooWrapper
中每个元素的类型?遵循以下原则:class FooWrapper<T> {
private Foo<T> foo;
public static FooWrapper<?> from(Foo<?> toWrap) {
Class<E> type = toWrap.getType(); // i know this is wrong
return new FooWrapper<type>(toWrap); // this is very wrong
}
private FooWrapper(Foo<T> foo) {
this.foo = foo;
}
}
最佳答案
您只需引入一个泛型即可稍微修改FooWrapper#from
:
public static <E> FooWrapper<E> from(Foo<E> toWrap) {
return new FooWrapper<E>(toWrap);
}