how to compare autoboxed integer in java

 
 

Guess the output of the following Java Program.

 
// file name: Main.java
public class Main {
	public static void main(String args[]) {
		Integer x = 400, y = 400;
		if (x == y)
			System.out.println("Same");
		else
			System.out.println("Not Same");
	}
}

Output:

Not Same

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

The output of the following program is a surprise from Java.

 
// file name: Main.java
public class Main {
	public static void main(String args[]) {
		Integer x = 40, y = 40;
		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

Predict the output of the following program.

class GFG
{
	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: 

 

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