如何将不兼容类型的对象流式传输到列表中？

Question

我的问题是：给定一个人员名单，返回所有学生。

这是我的课程：

人物类

public class Person {

}

学生班

public class Student extends Person {

}

方法

public static List<Student> findStudents(List<Person> list) {

    return list.stream()

            .filter(person -> person instanceof Student)

            .collect(Collectors.toList());

}

我收到编译错误：incompatible types: inference variable T has incompatible bounds

如何使用流返回列表中的所有学生而不会出现此错误。

Answer 1

return&nbsp;list.stream() &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;.filter(Student.class::isInstance) &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;.map(Student.class::cast) &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;.collect(Collectors.toList());它应该在那里进行强制转换，否则，它仍然是一个Stream<Person>.&nbsp;该instanceof检查不执行任何强制转换。Student.class::isInstance和Student.class::cast只是我的偏好，您可以分别选择p -> p instanceof Student和p -> (Student)p。

Answer 2

你需要一个演员：public static List<Student> findStudents(List<Person> list)&nbsp;{&nbsp; &nbsp; return list.stream()&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;.filter(person -> person instanceof Student)&nbsp;&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;.map(person -> (Student) person)&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;.collect(Collectors.toList());}

Answer 3

另一种选择。public static List<Student> findStudents(List<Person> list)&nbsp;{&nbsp; &nbsp; return list.stream()&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; .filter(s -> Student.class.equals(s.getClass()))&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; .map(Student.class::cast)&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; .collect(Collectors.toList());}