流和不同的操作

Stream and the distinct operation(流和不同的操作)

问题描述

我有以下代码:

class C
{
    String n;

    C(String n)
    {
        this.n = n;
    }

    public String getN() { return n; }

    @Override
    public boolean equals(Object obj)
    {
        return this.getN().equals(((C)obj).getN());
    }
 }

List<C> cc = Arrays.asList(new C("ONE"), new C("TWO"), new C("ONE"));

System.out.println(cc.parallelStream().distinct().count());

但我不明白为什么 distinct 返回 3 而不是 2.

but I don't understand why distinct returns 3 and not 2.

推荐答案

您还需要覆盖 C 类中的 hashCode 方法.例如:

You need to also override the hashCode method in class C. For example:

@Override
public int hashCode() {
    return n.hashCode();
}

当两个 C 对象相等时，它们的 hashCode 方法必须返回相同的值.

When two C objects are equal, their hashCode methods must return the same value.

接口Stream的API文档没有提到这一点，但是众所周知，如果你重写equals，你也应该重写hashCode.Object.equals() 的 API 文档提到了这一点:

The API documentation for interface Stream does not mention this, but it's well-known that if you override equals, you should also override hashCode. The API documentation for Object.equals() mentions this:

请注意，每当重写 hashCode 方法时，通常都需要重写该方法，以维护 hashCode 方法的一般约定，该约定声明等于对象必须具有相同的哈希码.

Note that it is generally necessary to override the hashCode method whenever this method is overridden, so as to maintain the general contract for the hashCode method, which states that equal objects must have equal hash codes.

显然，Stream.distinct() 确实使用了对象的哈希码，因为当你像我上面展示的那样实现它时，你会得到预期的结果:2.

Apparently, Stream.distinct() indeed uses the hash code of the objects, because when you implement it like I showed above, you get the expected result: 2.

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