guava-libraries:Objects.hashCode(Object [])碰撞是否安全？

Question

在查看覆盖的不同选项时hashCode(),我被引导到Objects.hashCode(Object[])Google的guava-libraries(javadoc)中.javadoc表示它委托给Arrays.hashCode(Object[]).在许多不同的对象类型中使用此方法是否安全？这不容易发生哈希冲突,或者这不仅仅是因为容器通常只包含一种类型的对象？

举个简单的例子,考虑以下类,

public class Student {
    private final String name;

    public Student(String name) {
        this.name = name;
    }

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

public class Teacher {
    private final String name;

    public Teacher(String name) {
        this.name = name;
    }

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

public class HashCodeDriver {
    public static void main(String[] args) {
        final String name = "moe";
        Student s = new Student(name);
        Teacher t = new Teacher(name);

        long studentHash = s.hashCode();
        long teacherHash = t.hashCode();
        System.out.println("studentHash=" + studentHash + " teacherHash=" + teacherHash);
        if(studentHash == teacherHash) {
            System.out.println("hash codes match");
        }
        else {
            System.out.println("hash codes don't match");
        }
    }
}

输出:

studentHash=108322 teacherHash=108322
hash codes match

对象是两种不同的类型,但生成相同的哈希码.这不是问题吗？我应该将类作为第一个参数传入以防止此冲突吗？例如,

public class Student {
    private final String name;

    public Student(String name) {
        this.name = name;
    }

    @Override
    public int hashCode() {
        return Objects.hashCode(Student.class, name);
    }
}

public class Teacher {
    private final String name;

    public Teacher(String name) {
        this.name = name;
    }

    @Override
    public int hashCode() {
        return Objects.hashCode(Teacher.class, name);
    }
}

这就是javadoc警告只为这个方法提供单个对象的原因吗？来自javadoc,

警告:提供单个对象时,返回的哈希码不等于该对象的哈希码.

Answer 1

当2种不同类型的2个不同对象具有相同的哈希码时,这不是问题.

希望,当你打算建立自己的时候,HashMap你不会把学生和教师作为该地图的关键.即使在你想做的情况下,HashMap<Object, Object>你也会没事,因为

assertFalse( new Teacher( "John Smith" ).equals( new Student( "John Smith" ) );

这就是为什么它同时重写是很重要的hashCode和equals.

委托的唯一缺点Arrays.hashCode(Object[])可能是,从性能的角度来看,有时可能太昂贵.

例如,在您的情况下,对于教师或学生来说,这将是一个更好的哈希方法.

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