As the name suggests, String Pool in java is a pool of Strings stored in Java Heap Memory. We know that String is a special class in java and we can create String objects using a new operator as well as providing values in double-quotes.
String Pool in Java
Here is a diagram that clearly explains how String Pool is maintained in java heap space and what happens when we use different ways to create Strings. 
String Pool is possible only because String is immutable in Java and its implementation of String interning concept. String pool is also example of Flyweight design pattern. String pool helps in saving a lot of space for Java Runtime although it takes more time to create the String. When we use double quotes to create a String, it first looks for String with the same value in the String pool, if found it just returns the reference else it creates a new String in the pool and then returns the reference. However using new operator, we force String class to create a new String object in heap space. We can use intern() method to put it into the pool or refer to another String object from the string pool having the same value. Here is the java program for the String Pool image:
package com.journaldev.util;
public class StringPool {
/**
* Java String Pool example
* @param args
*/
public static void main(String[] args) {
String s1 = "Cat";
String s2 = "Cat";
String s3 = new String("Cat");
System.out.println("s1 == s2 :"+(s1==s2));
System.out.println("s1 == s3 :"+(s1==s3));
}
}
Output of the above program is:
s1 == s2 :true
s1 == s3 :false
Recommended Read: Java String Class
How many Strings are getting Created in the String Pool?
Sometimes in java interview, you will be asked a question around String pool. For example, how many strings are getting created in the below statement;
String str = new String("Cat");
In the above statement, either 1 or 2 string will be created. If there is already a string literal “Cat” in the pool, then only one string “str” will be created in the pool. If there is no string literal “Cat” in the pool, then it will be first created in the pool and then in the heap space, so a total of 2 string objects will be created. Read: Java String Interview Questions
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hi … excellent post … one correction - “we force String class to create a new String object and then place it in the pool.” here, new operator creates object in Java heap, not in pool.
- abhay agarwal
String Pool only contains reference to the String objects. The text means the same thing.
- Pankaj
Yes, Pankaj. Please update this post accordingly. Your other posts also say so.
- Rishi Raj
Where the string stored.Means location of string where it stored.
- Ridham Shah
Could you post some example on intern() method.
- siddu
https://stackoverflow.com/questions/17489250/how-can-a-string-be-initialized-using/17489410#17489410
- Vikash
I have one confusion, when we create a new String Object like String s = new String (“abc”); As far as I know, this will create a new String Object in Heap and also abc will go in String constant pool and s will point value exists in string constant pool… which means the value already exists in String Constant Pool, so what will intern do in this case? Also String Constant Pool is not a part of Java Heap Memory…?
- kunal
When we use new keyword to create the String object, it doesn’t go into the String Pool. When we call intern method, then it checks if any String with same value is already present in the pool or not. If there is already a String with same value, it returns the reference or else it places the String into the pool.
- Pankaj
Hi Kunal, you can refer below snippet String str1 = “abc”; String str2 = “ab”; String str3 = “c”; String str4 = str2 + str3;//using StringBuilder for concatenation and returns new String object System.out.println(str1 == str4); // returns false str4 = str4.intern(); System.out.println(str1 == str4); // returns true
- Ganesh Rashinker
in frist question You are Right. when we create String by new keyword then one will go to heap and another in String pool String constant pool is depend on the JVM. architecture
- vinod
How come String Pool has two Cat literals and is Stringpool a part of Heap?
- Ganesh
Corrected the image, yes String Pool is part of Java Heap memory.
- Pankaj
String s = “ashish”; String b = new String(“ashish”); b.intern(); System.out.println(s==b); if string pool is manged then why not string ‘s’ reference is returned on b.intern() to b. It is printing false which means their reference are not same.
- Ashish
You need to write
b = b.intern();, see you haven’t updated the reference of “b” variable.- Pankaj
intern() method return a String . but in case of you does not hold return of intern ().
- vinod
If for example: String str1 = “abc”; String str2 = new String(“def”); Then, Case 1: String str3 = str1.concat(str2) will go in the heap or pool? Case 2: String str4 = str2.concat(“HI”) will go in the heap or pool?
- krishna
will be creating the new object in the string pool,
- krishna
public class MainDemo { public void comp() { String s1=“India”; String s2=“India”; String s3=new String(“India”); System.out.println(“== result:” + s1==s2); // false System.out.println(“equals result:” + s1.equals(s2)); // true System.out.println(“****************”); System.out.println(“== result:” + s1==s3); // false System.out.println(“equals result:” + s1.equals(s3)); // true } /** * @param args */ public static void main(String[] args) { // TODO Auto-generated method stub MainDemo obj=new MainDemo(); obj.comp(); } } Question - what is the difference between s1==s2 and s1==s3 , => both returns same result. but in your post s1==s2 is true that is not right result. I have checked this by writing same program in to this post. So plz clear my doubt ??
- sanjaya verma
“== result:” + s1==s2+ operator priority is more, so it will result in"== result:India"=="India"and hencefalse. I hope it will clear your confusion.- Pankaj
Thnx for the post. helped to clarify a lot of things
- balinder
In the name of god. public static void main(String[] args) { String s1 = “Cat”; String s2 = “Cat”; String s3 = new String(“Cat”); String test; System.out.println(“s1 == s2 :”+(s1==s2)); System.out.println(“s1 == s3 :”+(s1==s3)); } hi, how to return value(No address) from heap(s3) to stack(test). Thanks.
- Amir hossein
package com.journaldev.util; public class StringPool { /** * Java String Pool example * @param args */ public static void main(String[] args) { String s1 = “Cat”; String s2 = “Cat”; // here s1 & s2 value Cat stored into String Constant pool only once, so no doubt. String s3 = new String(“Cat”); String s4 = new String(“Cat”) ; // my question is here. here s3 & s4 value Cat is two times created into normal pool or only once created into normal pool. System.out.println(“s1 == s2 :”+(s1==s2)); System.out.println(“s1 == s3 :”+(s1==s3)); System.out.println(“s3 == s4 :”+(s3==s4)); // here both value of hashCode same but why to be false? } }
- Karuppasamy
Pankaj, Is there any way to get the size of String Pool?
- Pradeep Singh
String s=“abc”; s = s+“def”; System.out.println("String Result:- "+s);//output - String Result:- abcdef In this case content of ‘s’ variable is getting changed but according to java concept String is immutable. So,can you please explain me what happen with ‘s’ variable in the string pool?
- Pranita W
Thanks. But reply form is too far from article. I can forget my answer during scrolling down )
- Timofei
Hi , Can you please clarify me, how many objects will be created in below case and why? String s1=“cat”; String s2=“cat”; String s3=“c”+ “at”; String s4= new String(“cat”); String s5=“kitty” + “cat”; String s6= new String(kittycat); Thanks In Advance. :)
- Satish
Hi Pankaj, I am following you articles and finding it really precise. Really appreciate your efforts. But coming to this article, I have one doubts. The value in the String pool are “String” or reference to the String Object created in the Heap i.e. collection of references to String objects ?
- Ravi
Hi, I have one example which I found a little confusing. I understand that in context of String object type, by using NEW word, new object is created and when using “” in String initialization, String pool mechanizm is used. Why isn’t String pool used in following case? ******************* String myStr = “good”; char[] myCharArr = {‘g’, ‘o’, ‘o’, ‘d’ }; String newStr = “”; for(char ch : myCharArr){ newStr = newStr + ch; } boolean b1 = newStr == myStr; boolean b2 = newStr.equals(myStr); System.out.println(b1+ " " + b2); (false true is printed) ******************* Thx in advance, Teodor
- Teo
String s=“abc” ;-----creates a string object “abc” with reference ‘s’ in string pool. String s=new String(“abc”);----creates a string object “abc” with reference ‘s’ in heap memory and also creates a string object “abc” in string pool without reference so here two objects are created then what is the necessity for creating String object using new operator?
- muni
Another thing about the Java String Pool. For long term maintenance of your code you should use the “equals” method not the “==”. You may refactor the code and string pool object is replaced with a heap object and you get all kinds of problems. Refactor tools don’t “see” the cases where == should be changed to equal. primitive => == objects => equals
- Adrian