Difference Between String, StringBuilder And StringBuffer Classes in Java

Here we will explain the difference between String , StringBuilder and StringBuffer . As you will find that there are minor differences between the above mentioned classes.

String

• String is immutable( once created cannot be changed ) object.
• The object created as a String is stored in the Constant String Pool
• Every immutable object in Java is thread safe, that implies String is also thread safe.
• String cannot be used by two threads simultaneously.
• String once assigned cannot be changed.

Example

String  demo = " hello " ;

// The above object is stored in constant string pool and its value cannot be modified.

demo = "Bye" ;     //new "Bye" string is created in constant pool and referenced by the demo variable 

// "hello" string still exists in string constant pool and its value is not overridden but we lost reference to the "hello" string

StringBuffer 

• StringBuffer is mutable means one can change the value of the object .
• The object created through StringBuffer is stored in the heap . - StringBuffer has the same methods as the StringBuilder , but each method in StringBuffer is synchronized that is StringBuffer is thread safe.

Due to this it does not allow two threads to simultaneously access the same method. Each method can be accessed by one thread at a time.

But being thread safe has disadvantages too as the performance of the StringBuffer hits due to thread safe propert . Thus StringBuilder is faster than the StringBuffer when calling the same methods of each class.

String Buffer can be converted to the string by using toString() method.

Example:

StringBuffer demo1 = new StringBuffer("Hello") ;
// The above object stored in heap and its value can be changed.
demo1=new StringBuffer("Bye")
demo1=demo1.append("See You");
If you print demo1 you will get "HelloByeSee You"

StringBuilder 

• StringBuilder is same as the StringBuffer , i.e. the object is stored in heap and it can also be modified.
• The main difference between the StringBuffer and StringBuilder is that StringBuilder is also not thread safe. It have all the methods of StringBuffer - StringBuilder is fast as it is not thread safe.

	String	StringBuffer	StringBuilder
Storage	Constant String Pool	Heap	Heap
Mutable	No	Yes	Yes
Thread Safe	Yes	Yes	No
Performance	Fast	Slow	Fast

Hope you guys like it. Let me now in comments if you guys have any doubts. Happy Learning !!!

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Explain Strings in Java

In this we are going to learn about Strings in Java

• The String class is immutable (constant), i.e. Strings in java, once created and initialized, cannot be changed.
• The String is a final class, no other class can extend it, and you cannot change the state of the string.
• String values cannot be compare with '==', for string value comparision, use equals() method.
• String class supports various methods, including comparing strings, extracting substrings, searching characters & substrings, converting into either lower case or upper case, etc.
• The string examples given below describes more on string functionlaity.

The follwoing example expalin's about different way's of string initilization

package com.onlinecodegeek.javase;
public class StringInitializationEx {
    public static void main(String a[]){
        String str1 = "This is a string object";
        String str2 = new String("this is also string object");
        char[] c = {'a','b','c','d'};
        String str4 = new String(c);
        String str5 = str1+" This is another String object";
        System.out.println("str1"+str1);
        System.out.println("str2"+str2);
        System.out.println("str4"+str4);
        System.out.println("str5"+str5);
    }
}

Output

str1This is a string object str2this is also string object str4abcd str5This is a string object This is another String object

 

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How to find largest word in a given string using Java with O(n)

package com.onlinecodegeek.java;

public class FindLargetWord {

 public static void main(String args[]) {
  String s = " Welcome to Java World";
  String[] wordsArray = s.split(" ");
  int maxsize = 0;
  String maxWord = "";
  for (int i = 0; i < wordsArray.length; i++) {
   if (wordsArray[i].length() > maxsize) {
    maxWord = wordsArray[i];
    maxsize = wordsArray[i].length();
   }

  }
  System.out.println("Max sized word is  " + maxWord + " with  size " + maxsize);
 }
}

 

Thulasiram

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How to find Length of Last Word in long String

How to find Length of Last Word Length in long String

SOURCE : LastWordLength.java
INPUT

String s = "i love this country and and also it seems that we have the correctness of the somengdfer hereIam   ";

OUTPUT

7

Hi, I have written and developed this post so that most of people will be benefited. I'm committed to provide easy and in-depth tutorials on various technologies.I hope it will help you a lot.
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