Integer wrapper object interning

As you already know the == operator always checks the reference or objects. So, whenever you create an object of Integer class(Wrapper class), the valueOf() method called every time.

The value of the method checks the integer value if the value lies between -128 to 127 it returns the same reference. If value exceed from the limit then it creates a new object.
You can check the code of the ValueOf() method in JDK.

public static Integer valueOf(int i) {
        if (i >= IntegerCache.low && i <= IntegerCache.high)
            return IntegerCache.cache[i + (-IntegerCache.low)];
        return new Integer(i);


public class Program
    public static void main(String[] args) 
    	Integer a = 100;
    	Integer b = 100;
    	System.out.println(a == b);
    	System.out.println(a == b);

Output: true

6 thoughts on “Integer wrapper object interning”

    • Hi Anirudh Sharma,
      When we create an integer by direct assignment of an int literal. Then compiler converts it to an Integer reference.
      For example:
      Integer a = 100;
      In this example, the compiler makes conversion from int literal to reference by use of the auto-boxing concept.
      Integer a = Integer.valueOf(100);.

      The Integer class(Wrapper) has an internal IntegerCache and the range of cache from -128 to 127(By default). So whenever we create int literal the Integer.valueOf() method returns objects of mentioned range from that cache. So a == b returns true because a and b both are pointing to the same object.

      The valueOf() method checks the integer value. If the value of Integer is lies between -128 to 127 it returns the same reference from the pool. But if the value doesn’t lie in the limit it will create a new Object.

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