Java:为方便起见,在equals()中使用hashCode()？

Question

考虑以下测试用例,将equals中的hashCode()方法用作方便的快捷方式是不好的做法？

public class Test 
{    
    public static void main(String[] args){
        Test t1 = new Test(1, 2.0, 3, new Integer(4));
        Test t2 = new Test(1, 2.0, 3, new Integer(4));
        System.out.println(t1.hashCode() + "\r\n"+t2.hashCode());
        System.out.println("t1.equals(t2) ? "+ t1.equals(t2));
    }

    private int myInt;
    private double myDouble;
    private long myLong;
    private Integer myIntObj;

    public Test(int i, double d, long l, Integer intObj ){
        this.myInt = i;
        this.myDouble = d;
        this.myLong = l;
        this.myIntObj = intObj;
    }

    @Override
    public boolean equals(Object other)
    {        
        if(other == null) return false;
        if (getClass() != other.getClass()) return false;            

        return this.hashCode() == ((Test)other).hashCode();//Convenient shortcut?
    }

    @Override
    public int hashCode() {
        int hash = 3;
        hash = 53 * hash + this.myInt;
        hash = 53 * hash + (int) (Double.doubleToLongBits(this.myDouble) ^ (Double.doubleToLongBits(this.myDouble) >>> 32));
        hash = 53 * hash + (int) (this.myLong ^ (this.myLong >>> 32));
        hash = 53 * hash + (this.myIntObj != null ? this.myIntObj.hashCode() : 0);
        return hash;
    }   
}

主方法输出:

1097562307
1097562307
t1.equals(t2) ? true

Answer 1

通常,比较hashCode()而不是使用equals()并不安全.当equals()返回false时,hashCode()可以根据hashCode()的约定返回相同的值.

Answer 2

很坏!HashCode相等并不意味着equals返回true.合同是两个相等的对象必须具有相同的hashCode.但它没有说明具有相同HashCode的两个对象必须相等.

Answer 3

正如所有其他答案所述,这是不好的做法. 但是,您可能希望在equals方法中引用哈希代码的一种情况是,您有一个不可变对象并且先前已缓存哈希代码.这允许您在执行完整比较之前执行廉价的不精确比较.例如:

public class MyImmutable {
    private final String a;
    private final String b;
    private final int c;

    private int hashCode;

    public MyImmutable(String a, String b, int c) {
        this.a = a;
        this.b = b;
        this.c = c;
    }

    @Override
    public boolean equals(Object o) {
        if (this == o) return true;
        if (o == null || getClass() != o.getClass()) return false;

        MyImmutable that = (MyImmutable) o;

        // Compare cached hashCodes first before performing more expensive comparison.
        return hashCode == that.hashCode() && c == that.c && !(a != null ? !a.equals(that.a) : that.a != null) && !(b != null ? !b.equals(that.b) : that.b != null);
    }

    @Override
    public int hashCode() {
        if (hashCode == 0) {
            // hashCode not cached, or it was computed as 0 (we recalculate it in this case).
            hashCode = a != null ? a.hashCode() : 0;
            hashCode = 31 * hashCode + (b != null ? b.hashCode() : 0);
            hashCode = 31 * hashCode + c;
        }

        return hashCode;
    }
}

Answer 4

这不行.从本质上讲,哈希德斯不保证是独一无二的.

Answer 5

不好的做法？更重要的是,这是完全错误的.两个不相等的对象可以返回相同的哈希码.不要这样做.