[英]Should to objects which have the same hash be equal?
在下面的示例中,我創建了兩個具有完全相同內部結構的對象。 兩者都只攜帶值 1 作為實例變量。 我的想法是,如果我取e1
的哈希值,它應該與e2
的哈希值相同,因此e1.equals(e2)
應該返回 true。
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;
}
}
}
equals(Object)
的默認實現檢查兩個對象是否是同一個實例(即它們是==
)。 如果你想要一些不同的邏輯,你必須自己實現它。 請注意,如果您這樣做,您還應該實現自己的hashCode()
,以便兩個相等的對象也將具有匹配的哈希碼。 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;
}
}
從https://docs.oracle.com/javase/9/docs/api/java/lang/Object.html#hashCode看以下幾點——
現在,查看以下代碼及其輸出:
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());
}
}
輸出:
true
employee1.hashCode(): 511833308
employee3.hashCode(): 511833308
false
employee2.hashCode(): 1297685781
由於employee3
指向同一對象employee1
,你所得到的相同的哈希碼為他們而employee2
指向一個不同的對象(盡管它具有相同的內容,關鍵字, new
會在內存中創建一個單獨的對象),因此,您可能很少從文檔狀態中獲得與上面提到的第 4 點相同的employee2
哈希碼: As much as is reasonably practical, the hashCode method defined by class Object does return distinct integers for distinct objects.
您必須以某種方式覆蓋hashCode
方法,該方法應該為具有相同內容的兩個對象返回相同的哈希碼,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());
}
}
輸出:
true
employee1.hashCode(): 128107556
employee3.hashCode(): 128107556
false
employee2.hashCode(): 128107556
上面給出的hashCode
的實現為employee1
和employee2
生成相同的hashcode,即使equals
返回false
(檢查文檔中提到的第3 點)。
覆蓋hashCode
錯誤方式可能會導致即使相同的對象返回不同的哈希碼,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());
}
}
輸出:
true
employee1.hashCode(): 66066760
employee1.hashCode() again: 66069457
employee3.hashCode(): 66073797
false
employee2.hashCode(): 66074882
這是覆蓋hashCode
的錯誤方法,因為在 Java 應用程序執行期間多次調用同一個對象上的hashCode
必須始終如一地返回相同的整數(檢查文檔中提到的第 1 點)。
現在,查看以下代碼及其輸出:
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());
}
}
輸出:
true
employee1.hashCode(): 511833308
employee3.hashCode(): 511833308
true
employee2.hashCode(): 1297685781
由於employee1.equals(employee2)
返回true
,哈希碼也應該返回相同的(檢查點#2上面提到的文檔)。 但是, employee1
和employee2
的hashcode 值不同,這是不正確的。 這種差異是因為我們沒有覆蓋hashCode
方法。 因此,每當您覆蓋equals
,您還應該以正確的方式覆蓋hashCode
。
最后,下面給出了實現hashCode
和equals
的正確方法:
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());
}
}
輸出:
true
employee1.hashCode(): 128107556
employee3.hashCode(): 128107556
true
employee2.hashCode(): 128107556
您需要覆蓋equals
方法,否則將使用Object
的equals
方法來比較兩個實例。
@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;
}
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