具有相同散列值的 objects 应该相等吗？

Question

在下面的示例中，我创建了两个具有完全相同内部结构的 objects。两者都只携带值 1 作为实例变量。我的想法是，如果我采用 e1 的散列，它应该与 e2 的散列相同，因此 e1.equals(e2) 应该 return 为真。

class EqualsChecker {

    public static void main(String[] args) {

        Elem e1 = new Elem(1);
        Elem e2 = new Elem(1);


        System.out.println(e1);                                // EqualsChecker$Elem@6ff3c5b5
        System.out.println(e2);                                // EqualsChecker$Elem@3764951d
        System.out.println("e1.equals(e2): " + e1.equals(e2)); // returns false
    }


    static class Elem {
        private int v;
        public Elem(int i) {
            this.v = i;
        }   
    }   
}

这里为什么equals return false？我想我在下面的草图中有中间情况：

Answer 1

equals(Object) 的默认实现检查两个对象是否是同一个实例（即它们是 ==）。如果你想要一些不同的逻辑，你必须自己实现它。请注意，如果这样做，您还应该实现自己的 hashCode()，这样两个相等的对象也将具有匹配的哈希码。例如：

class Elem {
    private int v;

    @Override
    public boolean equals(final Object o) {
        if (o == null || this.getClass() != o.getClass()) {
            return false;
        }
        Elem elem = (Elem) o;
        return this.v == elem.v;
    }

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

Answer 2

从https://docs.oracle.com/javase/9/docs/api/java/lang/Object.html#hashCode--

看以下几点

只要在 Java 应用程序的执行过程中对同一个对象多次调用，hashCode 方法必须始终 return 相同的整数，前提是没有在 equals 比较中使用的信息对象被修改。从一个应用程序的一次执行到同一应用程序的另一次执行，该整数不需要保持一致。
如果根据 equals(Object) 方法两个对象相等，则对这两个对象中的每一个调用 hashCode 方法必须产生相同的整数结果。
不要求如果根据 equals(java.lang.Object) 方法两个对象不相等，则对两个对象中的每一个调用 hashCode 方法必须产生不同的整数结果。但是，程序员应该意识到，为不相等的对象生成不同的整数结果可能会提高哈希表的性能。
在相当实用的情况下，class Object 定义的 hashCode 方法为不同的对象执行 return 不同的整数。（hashCode 可能会或可能不会在某个时间点实现为对象内存地址的某些函数。）

现在，查看以下代码及其输出：

class MyEmployee {
    String code;
    String name;
    int age;

    public MyEmployee(String code, String name, int age) {
        super();
        this.code = code;
        this.name = name;
        this.age = age;
    }
}

public class Main {
    public static void main(String[] args) {
        MyEmployee employee1 = new MyEmployee("AB12", "Dhruv", 24);
        MyEmployee employee2 = new MyEmployee("AB12", "Dhruv", 24);
        MyEmployee employee3 = employee1;
        System.out.println(employee1.equals(employee3));
        System.out.println("employee1.hashCode(): " + employee1.hashCode());
        System.out.println("employee3.hashCode(): " + employee3.hashCode());
        System.out.println(employee1.equals(employee2));
        System.out.println("employee2.hashCode(): " + employee2.hashCode());
    }
}

输出：

true
employee1.hashCode(): 511833308
employee3.hashCode(): 511833308
false
employee2.hashCode(): 1297685781

由于 employee3 指向与 employee1 相同的对象，因此您将获得相同的哈希码，而 employee2 指向不同的对象（尽管它们具有相同的哈希码）内容，关键字，new 将在内存中创建一个单独的对象），因此，您可能很少获得 employee2 与提到的 point#4 相同的哈希码以上来自文档说明：As much as is reasonably practical, the hashCode method defined by class Object does return distinct integers for distinct objects.

您必须以某种方式重写 hashCode 方法，使具有相同内容的两个对象 return 具有相同的哈希码，例如

class MyEmployee {
    String code;
    String name;
    int age;

    public MyEmployee(String code, String name, int age) {
        super();
        this.code = code;
        this.name = name;
        this.age = age;
    }

    @Override
    public int hashCode() {
        final int prime = 31;
        int result = 1;
        result = prime * result + age;
        result = prime * result + ((code == null) ? 0 : code.hashCode());
        result = prime * result + ((name == null) ? 0 : name.hashCode());
        return result;
    }   
}

public class Main {
    public static void main(String[] args) {
        MyEmployee employee1 = new MyEmployee("AB12", "Dhruv", 24);
        MyEmployee employee2 = new MyEmployee("AB12", "Dhruv", 24);
        MyEmployee employee3 = employee1;
        System.out.println(employee1.equals(employee3));
        System.out.println("employee1.hashCode(): " + employee1.hashCode());
        System.out.println("employee3.hashCode(): " + employee3.hashCode());
        System.out.println(employee1.equals(employee2));
        System.out.println("employee2.hashCode(): " + employee2.hashCode());
    }
}

输出：

true
employee1.hashCode(): 128107556
employee3.hashCode(): 128107556
false
employee2.hashCode(): 128107556

上面给出的 hashCode 的实现为 employee1 和 employee2 生成相同的哈希码，即使 equals returns false (检查上面从文档中提到的 point#3。

覆盖 hashCode 的错误方法甚至可能导致相同的对象 return 使用不同的哈希码，例如

class MyEmployee {
    String code;
    String name;
    int age;

    public MyEmployee(String code, String name, int age) {
        super();
        this.code = code;
        this.name = name;
        this.age = age;
    }

    @Override
    public int hashCode() {
        final int prime = 31;
        int result = 1;
        result = prime * result + age;
        result = prime * result + ((code == null) ? 0 : (int) (code.length() * (Math.random() * 100)));
        result = prime * result + ((name == null) ? 0 : name.hashCode());
        return result;
    }
}

public class Main {
    public static void main(String[] args) {
        MyEmployee employee1 = new MyEmployee("AB12", "Dhruv", 24);
        MyEmployee employee2 = new MyEmployee("AB12", "Dhruv", 24);
        MyEmployee employee3 = employee1;
        System.out.println(employee1.equals(employee3));
        System.out.println("employee1.hashCode(): " + employee1.hashCode());
        System.out.println("employee1.hashCode() again: " + employee1.hashCode());
        System.out.println("employee3.hashCode(): " + employee3.hashCode());
        System.out.println(employee1.equals(employee2));
        System.out.println("employee2.hashCode(): " + employee2.hashCode());
    }
}

输出：

true
employee1.hashCode(): 66066760
employee1.hashCode() again: 66069457
employee3.hashCode(): 66073797
false
employee2.hashCode(): 66074882

这是覆盖 hashCode 的错误方法，因为在执行 Java 应用程序期间多次调用同一对象 hashCode 必须始终 return相同的整数（检查上面文档中提到的 point#1）。

现在，查看以下代码及其输出：

class MyEmployee {
    String code;
    String name;
    int age;

    public MyEmployee(String code, String name, int age) {
        super();
        this.code = code;
        this.name = name;
        this.age = age;
    }

    @Override
    public boolean equals(Object obj) {
        if (this == obj)
            return true;
        if (obj == null)
            return false;
        if (getClass() != obj.getClass())
            return false;
        MyEmployee other = (MyEmployee) obj;
        if (code == null) {
            if (other.code != null)
                return false;
        } else if (!code.equals(other.code))
            return false;
        return true;
    }
}

public class Main {
    public static void main(String[] args) {
        MyEmployee employee1 = new MyEmployee("AB12", "Dhruv", 24);
        MyEmployee employee2 = new MyEmployee("AB12", "Dhruv", 24);
        MyEmployee employee3 = employee1;
        System.out.println(employee1.equals(employee3));
        System.out.println("employee1.hashCode(): " + employee1.hashCode());
        System.out.println("employee3.hashCode(): " + employee3.hashCode());
        System.out.println(employee1.equals(employee2));
        System.out.println("employee2.hashCode(): " + employee2.hashCode());
    }
}

输出：

true
employee1.hashCode(): 511833308
employee3.hashCode(): 511833308
true
employee2.hashCode(): 1297685781

因为 employee1.equals(employee2) returns true，哈希码也应该 returned 相同（检查 point#2上面从文档中提到的）。但是，employee1 和 employee2 的哈希码值不同，这是不正确的。这种差异是因为我们没有覆盖 hashCode 方法。因此，无论何时覆盖 equals，您也应该以正确的方式覆盖 hashCode。

最后，下面给出了 hashCode 和 equals 的正确实现方式：

class MyEmployee {
    String code;
    String name;
    int age;

    public MyEmployee(String code, String name, int age) {
        super();
        this.code = code;
        this.name = name;
        this.age = age;
    }

    @Override
    public int hashCode() {
        final int prime = 31;
        int result = 1;
        result = prime * result + age;
        result = prime * result + ((code == null) ? 0 : code.hashCode());
        result = prime * result + ((name == null) ? 0 : name.hashCode());
        return result;
    }

    @Override
    public boolean equals(Object obj) {
        if (this == obj)
            return true;
        if (obj == null)
            return false;
        if (getClass() != obj.getClass())
            return false;
        MyEmployee other = (MyEmployee) obj;
        if (age != other.age)
            return false;
        if (code == null) {
            if (other.code != null)
                return false;
        } else if (!code.equals(other.code))
            return false;
        if (name == null) {
            if (other.name != null)
                return false;
        } else if (!name.equals(other.name))
            return false;
        return true;
    }
}

public class Main {
    public static void main(String[] args) {
        MyEmployee employee1 = new MyEmployee("AB12", "Dhruv", 24);
        MyEmployee employee2 = new MyEmployee("AB12", "Dhruv", 24);
        MyEmployee employee3 = employee1;
        System.out.println(employee1.equals(employee3));
        System.out.println("employee1.hashCode(): " + employee1.hashCode());
        System.out.println("employee3.hashCode(): " + employee3.hashCode());
        System.out.println(employee1.equals(employee2));
        System.out.println("employee2.hashCode(): " + employee2.hashCode());
    }
}

输出：

true
employee1.hashCode(): 128107556
employee3.hashCode(): 128107556
true
employee2.hashCode(): 128107556

Answer 3

您需要重写equals方法，否则Object的equals方法将用于比较两个实例。

@Override
public boolean equals(Object that) {
    if (this == that) return true;
    if (that instanceof Elem) {
        Elem thatElem = (Elem) that;
        return thatElem.v == this.v;
    }
    return false;
}

具有相同散列值的 objects 应该相等吗？

Should to objects which have the same hash be equal?

java

equality