How is creation of a String with new() different from a literal?

In this shot, we will discuss how the creation of a String using the new() method is different from that of a literal in Java.

The table and illustration below show the difference between String creation using the new() method and String creation using a String literal.

String creation using new()

String creation using String literal

If we create a String using new(), then a new object is created in the heap memory even if that value is already present in the heap memory.

If we create a String using String literal and its value already exists in the string pool, then that String variable also points to that same value in the String pool without the creation of a new String with that value.

It takes more time for the execution and thus has lower performance than using String literal.

It takes less time for the execution and thus has better performance than using new().

Example:

String n1= new String(“Java”);

String n2= new String(“Java”);

String n3= new String(“Create”);

Example:

String s1=”Java”;

String s2=”Java”;

String s3=”Create”;

Code

The code below approaches this problem by creating String objects through both methods. We use the == operator for reference comparison and justify the difference between the objects.

class Main
{
    public static void main(String[] args)
    {
        String s1="Java";
        String s2="Java";
        String s3="Create";
        String n1= new String("Java");
        String n2= new String("Java");
        String n3= new String("Create");
        System.out.println(s1==s2);
        System.out.println(n1==n2);
        System.out.println(s3==n3);
    }
}

Explanation

  • In line 1, we create a Main class with a main function.

  • In lines 5 to 7, we use String literals to initialize three String type variables.

  • In lines 9 to 11, we use new() to initialize three String objects.

  • In line 13, we compare the references of s1 and s2, which point to the same objects, and therefore display true.

  • In line 14, we compare the references of n1 and n2, which do not point to the same objects, and therefore display false.

  • In line 15, we compare the references of s3 and n3, which do not point to the same objects, and display false.

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