你們在使用equals方法時,一個常見的錯誤根源就是重寫了equals方法,而沒有覆蓋hashCode方法。java
在每一個覆蓋equals方法的類中,也必須覆蓋hashCode方法。若是不覆蓋的話,就會違反Object.hashCode的通用約定,從而致使該類沒法正常工做。ide
下面看個實例:this
public final class PhoneNumber { private final short areaCode; private final short prefix; private final short lineNumber; public PhoneNumber(int areaCode, int prefix, int lineNumber) { rangeCheck(areaCode, 999, "area code"); rangeCheck(prefix, 999, "prefix"); rangeCheck(lineNumber, 9999, "line number"); this.areaCode = (short) areaCode; this.prefix = (short) prefix; this.lineNumber = (short) lineNumber; } private static void rangeCheck(int arg, int max, String name) { if (arg < 0 || arg > max) throw new IllegalArgumentException(name + ": " + arg); } @Override public boolean equals(Object o) { if (o == this) return true; if (!(o instanceof PhoneNumber)) return false; PhoneNumber pn = (PhoneNumber) o; return pn.lineNumber == lineNumber && pn.prefix == prefix && pn.areaCode == areaCode; } public static void main(String[] args) { Map<PhoneNumber, String> m = new HashMap<PhoneNumber, String>(); m.put(new PhoneNumber(707, 867, 5309), "Jenny"); System.out.println(m.get(new PhoneNumber(707, 867, 5309))); }
這時候,可能也有不少人認爲輸出"Jenny",但它實際卻返回null。因爲沒有覆蓋hashCode方法,從而致使兩個相等的實例具備不相等的散列碼,違反了hashCode的約定。spa
修正這個問題也很簡單,只須要爲該類提供想對應的hashCode方法便可。code
因此,在每一個覆蓋equals方法的類中,也必須覆蓋hashCode方法。get