What is the difference between string object and string literal

+1 vote
Feb 28, 2019 in Java by Phalguni
• 1,020 points
67,367 views

4 answers to this question.

+1 vote

Difference between string objects and string literals

A string literal in Java is basically a sequence of characters from the source character set used by Java programmers to populate string objects or to display text to a user. These characters could be anything like letters, numbers or symbols which are enclosed within two quotation marks.

A string object is one which lets you work with a series of characters. It basically wraps all Javascript’s string primitive data type which a number of helper methods. The main function of JavaScript is it automatically converts between string primitives and string objects.

The main difference between String objects and string literals is mentioned below:

String Objects

String Literals

A new keyword is used to create an object and the object is created in the heap memory whereas its reference will be pointed to String pool.

Here the reference will be directly referred to String Pool. JVM initially checks for the availability with the same value in the constant pool

Here is an illustration of a Java Program that compares their performances.

  

class ComparePerformance {
      public static void main(String args[])
    {    
        long start1 = System.currentTimeMillis();
                  for (int i = 0; i < 10000; i++)
        {
            String s1 = "Hello World";
            String s2 = "Welcome";
        }
                 long end1 = System.currentTimeMillis();
        long total_time = end1 - start1;
          System.out.println("Time taken to execute"+ 
                " string literal = " + total_time);
                long start2 = System.currentTimeMillis();
                 for (int i = 0; i < 10000; i++)
        {
            String s3 = new String("Hello World");
            String s4 = new String("Welcome ");
        }
              long end2 = System.currentTimeMillis();
        long total_time1 = end2 - start2;
          System.out.println("Time taken to execute"+
                   " string object=" + total_time1);
    }
}
answered Feb 28, 2019 by Avantika
• 1,520 points

edited Mar 15, 2019 by Kalgi
+4 votes

String literal is a Java language concept. This is a String literal:

"a String literal"

String object is an individual instance of the java.lang.String class.

String s1 = "abcde";
String s2 = new String("abcde");
String s3 = "abcde";

All are valid, but have a slight difference. s1 will refer to an interned String object. This means, that the character sequence "abcde" will be stored at a central place, and whenever the same literal "abcde" is used again, the JVM will not create a new String object but use the reference of the cached String.

s2 is guranteed to be a new String object, so in this case we have:

s1 == s2 // is false
s1 == s3 // is true
s1.equals(s2) // is true
answered Aug 16, 2019 by Sirajul
• 59,230 points
and if you do

String s1 = new String("abc");

String s2 = new String("abc");

System.out.println(s1==s2)

op:

false
+1 vote

When you use a string literal the string can be interned, but when you use a new String("...") you get a new string object. In general, you should use the string literal notation when possible. It is easier to read and it gives the compiler a chance to optimize your code.

answered Dec 15, 2020 by Gitika
• 65,850 points
+1 vote

String Literal

String str = “java”;

This is string literal. When you declare string like this, you are actually calling intern() method on String. This method references internal pool of string objects. If there already exists a string value “Edureka”, then str will reference of that string and no new String object will be created. Please refer Initialize and Compare Strings in Java for details.

String Object

String str = new String(“Edureka”);

This is a string object. In this method, JVM is forced to create a new string reference, even if “Edureka” is in the reference pool.

Therefore, if we compare the performance of string literal and string object, a string object will always take more time to execute than a string literal because it will construct a new string every time it is executed.
Note: Execution time is compiler dependent.

Below is the Java program to compare their performances.

filter_none

edit

play_arrow

brightness_4
// Java program to compare performance 

// of string literal and string object

  

class ComparePerformance {

  

    public static void main(String args[])

    {    

        // Initialization time for String

        // Literal

        long start1 = System.currentTimeMillis();

          

        for (int i = 0; i < 10000; i++)

        {

            String s1 = "Edureka";

            String s2 = "Welcome";

        }

          

        long end1 = System.currentTimeMillis();

        long total_time = end1 - start1;

  

        System.out.println("Time taken to execute"+ 

                " string literal = " + total_time);

  

        // Initialization time for String

        // object

        long start2 = System.currentTimeMillis();

          

        for (int i = 0; i < 10000; i++)

        {

            String s3 = new String("Edureka");

            String s4 = new String("Welcome");

        }

          

        long end2 = System.currentTimeMillis();

        long total_time1 = end2 - start2;

  

        System.out.println("Time taken to execute"+

                   " string object=" + total_time1);

    }

}

Output:

Time is taken to execute string literal = 0
Time taken to execute string object = 2

answered Dec 15, 2020 by Roshni
• 10,520 points

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