【发布时间】：2019-03-16 19:48:30
【问题描述】：
假设我的 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);
}
}
上述程序的输出为：
[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employee类的equals()和hashcode()方法，两个对象是一样的。
我观察到，当我调用 distinct() 方法 equals() 和 hashcode() 方法时，不会得到 Set implementation 流的调用，但它致电获取列表实施流。
按照 JavaDoc 的说法 distinct() 返回由该流的不同元素（根据 Object.equals(Object)） 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();
【问题讨论】：
-
@MirkoAlicastro 您可以比较原语（例如
int，这是eno的类型），它不会通过引用而是通过值来比较它们。 -
@MirkoAlicastro 我给出的上述示例仅用于测试目的。我的问题是为什么 hashCode 和 equal 方法不会调用 If 我调用 distinct () 方法。
-
我看错了，我以为他们都是员工对象
-
即使是陌生人，
System.out.println(e1.equals(e2));在Stream部分之前返回true -
答案是你做错了操作：你正在编辑一个对象，插入一个哈希图中。插入对象时，不应修改哈希码使用的字段。您应该删除它，修改它，然后重新插入它，以便在 hashmap 中给它正确的位置