让我说我的Employee类具有正确覆盖的equals和hashcode方法。
public class Employee {
private int eno;
private String firstName;
private String lastName;
@Override
public int hashCode() {
System.out.println("hashcode called");
final int prime = 31;
int result = 1;
result = prime * result + eno;
result = prime * result + ((firstName == null) ? 0 : firstName.hashCode());
result = prime * result + ((lastName == null) ? 0 : lastName.hashCode());
return result;
}
@Override
public boolean equals(Object obj) {
System.out.println("equals called");
if (this == obj)
return true;
if (obj == null)
return false;
if (getClass() != obj.getClass())
return false;
Employee other = (Employee) obj;
if (eno != other.eno)
return false;
if (firstName == null) {
if (other.firstName != null)
return false;
} else if (!firstName.equals(other.firstName))
return false;
if (lastName == null) {
if (other.lastName != null)
return false;
} else if (!lastName.equals(other.lastName))
return false;
return true;
}
}
测试类如下
class Test {
public static void main(String[] args) {
Employee e1 = new Employee(1, "Karan", "Mehara");
Employee e2 = new Employee(2, "Rajesh", "Shukla");
Set<Employee> emps= new HashSet<>();
emps.add(e1);
emps.add(e2);
System.out.println(emps);
// No such requirement just for testing purpose modifying
e2.setEno(1);
e2.setFirstName("Karan");
e2.setLastName("Mehara");
System.out.println(emps);
emps.stream().distinct().forEach(System.out::println);
}
}
以上程序的输出为:
[员工[eno = 1,firstName = Karan,lastName = Mehara],员工[eno = 2,firstName = Rajesh,lastName = Shukla]]
[员工[eno = 1,firstName = Karan,lastName = Mehara],员工[eno = 1,firstName = Karan,lastName = Mehara]]
员工[eno = 1,firstName = Karan,lastName = Mehara]
员工[eno = 1,firstName = Karan,lastName = Mehara]
为什么distinct()方法返回重复元素?
根据雇员类的equals()和hashcode()方法,两个对象相同。
我观察到,当我调用distinct()方法equals()和hashcode()方法不会得到调用时,会流 Set实现,但是拨打电话以获取列表实现的流。
按照JavaDoc的说法 exclude()返回由此流的(根据Object.equals(Object))的不同元素组成的流。
/**
* Returns a stream consisting of the distinct elements (according to
* {@link Object#equals(Object)}) of this stream.
*
* <p>For ordered streams, the selection of distinct elements is stable
* (for duplicated elements, the element appearing first in the encounter
* order is preserved.) For unordered streams, no stability guarantees
* are made.
*
* <p>This is a <a href="package-summary.html#StreamOps">stateful
* intermediate operation</a>.
*
* @apiNote
* Preserving stability for {@code distinct()} in parallel pipelines is
* relatively expensive (requires that the operation act as a full barrier,
* with substantial buffering overhead), and stability is often not needed.
* Using an unordered stream source (such as {@link #generate(Supplier)})
* or removing the ordering constraint with {@link #unordered()} may result
* in significantly more efficient execution for {@code distinct()} in parallel
* pipelines, if the semantics of your situation permit. If consistency
* with encounter order is required, and you are experiencing poor performance
* or memory utilization with {@code distinct()} in parallel pipelines,
* switching to sequential execution with {@link #sequential()} may improve
* performance.
*
* @return the new stream
*/
Stream<T> distinct();
答案 0 :(得分:6)
Set
是defined,是“不包含重复元素的集合”。因此,Stream
的{{1}}的{{1}}方法很可能根本不执行任何操作,因为它已经确保值是唯一的。
Javadoc中明确提到了您所做的事情:
注意:如果将可变对象用作集合元素,则必须格外小心。如果对象的值更改为影响相等比较的方式,而该对象是集合中的元素,则不指定集合的行为。此禁止的一种特殊情况是,不允许集合将自身包含为元素。