Comparison of Autoboxed Integer objects in Java

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Comparison of Autoboxed Integer objects in Java

Comparison of Autoboxed Integer objects in Java

When we assign an integer value to an Integer object, the value is autoboxed into an Integer object. For example the statement “Integer x = 10” creates an object ‘x’ with value 10.

Following are some interesting output questions based on comparison of Autoboxed Integer objects.

Predict the output of following Java Program

// file name: Main.java
public class Main {
public static void main(String args[]) {
Integer a = 500, b = 500;
if (a == b)
System.out.println("Same");
else
System.out.println("Not Same");
}
}
Output:

Not Same

Since a and b refer to different objects, we get the output as “Not Same”

The output of following program is a surprise from Java.

// file name: Main.java
public class Main {
public static void main(String args[]) {
Integer x = 20, y = 20;
if (x == y)
System.out.println("Same");
else
System.out.println("Not Same");
}
}
Output:

Same

In Java, values from -128 to 127 are cached, so the same objects are returned. The implementation of valueOf() uses cached objects if the value is between -128 to 127.

If we explicitly create Integer objects using new operator, we get the output as “Not Same”. See the following Java program. In the following program, valueOf() is not used.

// file name: Main.java
public class Main {
public static void main(String args[]) {
Integer x = new Integer(40), y = new Integer(40);
if (x == y)
System.out.println("Same");
else
System.out.println("Not Same");
}
}
Output:

Not Same

class P_AI
{
public static void main(String[] args)
{
Integer X = new Integer(10);
Integer Y = 10;

// Due to auto-boxing, a new Wrapper object
// is created which is pointed by Y
System.out.println(X == Y);
}
}
Output:

false

Explanation: Two objects will be created here. First object which is pointed by X due to calling of new operator and second object will be created because of Auto-boxing.

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