distinct（）方法不为HashSet元素流返回不同的元素

Question

假设我有正确覆盖的equals和hashcode方法的Employee类。

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);
    }

}

上述程序的输出是：

[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，两个对象都是相同的。

我观察到，当我调用distinct（）方法时，equals（）和hashcode（）方法将不会调用Set实现流，但它会调用List实现流。

根据JavaDoc，distinct（）返回由此流的不同元素（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();

Answer 1

Set是defined是“不包含任何重复元素的集合”。因此，Stream的distinct-Set方法最有可能实现什么都不做，因为它已经证明价值是独一无二的。

您在Javadoc中明确提到了您所做的事情：

注意：如果将可变对象用作set元素，则必须非常小心。如果在对象是集合中的元素的同时以影响等于比较的方式更改对象的值，则不指定集合的​​行为。这种禁止的一个特例是，不允许集合将自身作为一个要素包含在内。