Effective Java 第八条：覆盖equals请遵守通用约定

Posted 2022-12-11 阿蛮家

首先介绍一下Java中equals通用约定，分别是：自反性、对称性、一致性、传递性。

检查自反性的通用方法：

将对象扔到List里，然后看List.contains()是否返回true。

首先来看一个违反对称性的例子：

public class CaseInsesitiveString 
    private final String s;

    public CaseInsesitiveString(String s) 
        if(s == null)
            throw new NullPointerException();
        
        this.s = s;
    

    @Override
    public boolean equals(Object obj) 
        if(obj instanceof CaseInsesitiveString)
            return s.equalsIgnoreCase(((CaseInsesitiveString) obj).s);
        

        if(obj instanceof String)
            return s.equalsIgnoreCase((String) obj);
        
        return false;
    

测试打印：

public class EqualTest 
    public static void main(String[] args) 
        CaseInsesitiveString str = new CaseInsesitiveString("Hello");
        String s = "hello";
        System.out.println(str.equals(s));
        System.out.println(s.equals(str));

        List<CaseInsesitiveString> list = new ArrayList<CaseInsesitiveString>();
        list.add(str);
        System.out.println(list.contains(s));
    

结果是：str.equals(s) 返回true,s.equals(str)返回false。显然违背了对称性。

原因是：String类中的equals并不知道大小写是否相等。

解决办法：

@Override
    public boolean equals(Object obj) 
        return obj instanceof CaseInsesitiveString 
                && s.equalsIgnoreCase(((CaseInsesitiveString) obj).s);
    

再来看一个违反传递性的例子：

public class ColorPoint extends Point 

    private final Color color;

    public ColorPoint(int x, int y, Color color) 
        super(x, y);
        this.color = color;
    

    @Override
    public boolean equals(Object obj) 
        if(!(obj instanceof Point))
            return false;
        if(!(obj instanceof ColorPoint))
            return obj.equals(this);
        return super.equals(obj) && ((ColorPoint) obj).color == color;
    

测试打印：

ColorPoint point = new ColorPoint(1,2,Color.BLUE);
        Point p = new Point(1,2);
        ColorPoint colorPoint = new ColorPoint(1,2, Color.RED);
        System.out.println(point.equals(p));
        System.out.println(p.equals(colorPoint));
        System.out.println(point.equals(colorPoint));

结果是：前面两个返回true，后面一个返回false。显然违反了传递性。

解决办法：

public class ColorPointNew 

    private final Point point;
    private final Color color;

    public ColorPointNew(int x, int y, Color color) 
        if(color == null)
            throw new NullPointerException();
        
        this.color = color;
        point = new Point(x,y);
    

    @Override
    public boolean equals(Object obj) 
        if(!(obj instanceof ColorPointNew))
            return false;
        ColorPointNew point = (ColorPointNew) obj;
        return point.point.equals(point) &&
                point.color == color;
    

    public Point asPoint()
        return point;