why 2 objects of Integer class in Java cannot be e

问题:

my code is:

public class Box
{
  public static void main(String[] args)
  {
     Integer z = new Integer(43);
     z++;

     Integer h = new Integer(44);

     System.out.println("z == h -> " + (h == z ));
  }
}

Output:-

z == h -> false

why the output is false when the values of both the objects is equal?

Is there any other way in which we can make the objects equal?

回答1:

You are trying to compare two different object and not their values. z and h points to two different Integer object which hold same value.

z == h 

Will check if two objects are equal. So it will return false.

If you want to compare values stored by Integer object use equals method.

---

Integer z = new Integer(43);   // Object1  is created with value as 43.
z++;                           // Now object1 holds 44.

Integer h = new Integer(44); // Object2 is created with value as 44.

So at the end we have two different Integer object ie object1 and object2 with value as 44. Now

z = h

This will check if objects pointed by z and h is same. ie object1 == object2 which is false. If you do

Integer z = new Integer(43);   // Object1  is created with value as 43.
z++;                           // Now object1 holds 44. Z pointing to Object1

Integer h = z;                 // Now h is pointing to same object as z.

Now

z == h  

will return true.

This might help http://www.programmerinterview.com/index.php/java-questions/java-whats-the-difference-between-equals-and/

回答2:

No. Use h.equals(z) instead of h == z to get the equality behavior you expect.

回答3:

h == z would work only if you assign the value via auto-boxing (i.e Integer a = 43) and the value is in between -128 and 127 (cached values), i.e:

 Integer a = 44;
 Integer b = 44;

 System.out.println("a == b -> " + (a == b));

OUTPUT:

a == b -> true

---

If the value is out of the range [-128, 127], then it returns false

 Integer a = 1000;
 Integer b = 1000;

 System.out.println("a == b -> " + (a == b));

OUTPUT:

a == b -> false

---

However, the right way to compare two objects is to use Integer.equals() method.

回答4:

Integer is Object not primitive (int) And Object equality compare with equals method.

When you do z == h it will not unbox into int value unless it checks both Integer reference(z & h) are referring same reference or not.

As it is derived in documentation -

The result is true if and only if the argument is not null and is an Integer object that contains the same int value as this object.

System.out.println("z == h -> " + h.equals( z));

It will print true.

回答5:

Integer is an object, not a primitive. If z & h were primitives, == would work just fine. When dealing with objects, the == operator doesn't check for equality; it checks if the two references point to the same object.

As such, use z.equals(h); or h.equals(z); These should return true.

回答6:

Read this: http://docs.oracle.com/javase/1.5.0/docs/api/java/lang/Integer.html

Integer is Object you compare address/references/pointers of objects not values.

Integer a = Integer(1);
Integer b = Integer(1);

a == b; // false
a.compareTo(b); // true

回答7:

check to make sure you can use z++ on the Integer object.