Vitamin C

Java Language Strings


Strings (java.lang.String) are pieces of text stored in your program. Strings are not a primitive data type in Java, however, they are very common in Java programs.

In Java, Strings are immutable, meaning that they cannot be changed. (Click here for a more thorough explanation of immutability.)


Since Java strings are immutable, all methods which manipulate a String will return a new String object. They do not change the original String. This includes to substring and replacement methods that C and C++ programers would expect to mutate the target String object.

Use a StringBuilder instead of String if you want to concatenate more than two String objects whose values cannot be determined at compile-time. This technique is more performant than creating new String objects and concatenating them because StringBuilder is mutable.

StringBuffer can also be used to concatenate String objects. However, this class is less performant because it is designed to be thread-safe, and acquires a mutex before each operation. Since you almost never need thread-safety when concatenating strings, it is best to use StringBuilder.

If you can express a string concatenation as a single expression, then it is better to use the + operator. The Java compiler will convert an expression containing + concatenations into an efficient sequence of operations using either String.concat(...) or StringBuilder. The advice to use StringBuilder explicitly only applies when the concatenation involves a multiple expressions.

Don't store sensitive information in strings. If someone is able to obtain a memory dump of your running application, then they will be able to find all of the existing String objects and read their contents. This includes String objects that are unreachable and are awaiting garbage collection. If this is a concern, you will need to wipe sensitive string data as soon as you are done with it. You cannot do this with String objects since they are immutable. Therefore, it is advisable to use a char[] objects to hold sensitive character data, and wipe them (e.g. overwrite them with '\000' characters) when you are done.

All String instances are created on the heap, even instances that correspond to string literals. The special thing about string literals is that the JVM ensures that all literals that are equal (i.e. that consists of the same characters) are represented by a single String object (this behavior is specified in JLS). This is implemented by JVM class loaders. When a class loader loads a class, it scans for string literals that are used in the class definition, each time it sees one, it checks if there is already a record in the string pool for this literal (using the literal as a key). If there is already an entry for the literal, the reference to a String instance stored as the pair for that literal is used. Otherwise, a new String instance is created and a reference to the instance is stored for the literal (used as a key) in the string pool. (Also see string interning).

The string pool is held in the Java heap, and is subject to normal garbage collection.

Java SE 7

In releases of Java before Java 7, the string pool was held in a special part of the heap known as "PermGen". This part was only collected occasionally.

Java SE 7

In Java 7, the string pool was moved off from "PermGen".

Note that string literals are implicitly reachable from any method that uses them. This means that the corresponding String objects can only be garbage collected if the code itself is garbage collected.

Up until Java 8, String objects are implemented as a UTF-16 char array (2 bytes per char). There is a proposal in Java 9 to implement String as a byte array with an encoding flag field to note if the string is encoded as bytes (LATIN-1) or chars (UTF-16).

Comparing Strings

In order to compare Strings for equality, you should use the String object's equals or equalsIgnoreCase methods.

For example, the following snippet will determine if the two instances of String are equal on all characters:

String firstString = "Test123";
String secondString = "Test" + 123;

if (firstString.equals(secondString)) {
   // Both Strings have the same content.

Live demo

This example will compare them, independent of their case:

String firstString = "Test123";
String secondString = "TEST123";

if (firstString.equalsIgnoreCase(secondString)) {
    // Both Strings are equal, ignoring the case of the individual characters.

Live demo

Note that equalsIgnoreCase does not let you specify a Locale. For instance, if you compare the two words "Taki" and "TAKI" in English they are equal; however, in Turkish they are different (in Turkish, the lowercase I is ı). For cases like this, converting both strings to lowercase (or uppercase) with Locale and then comparing with equals is the solution.

String firstString = "Taki";
String secondString = "TAKI";

System.out.println(firstString.equalsIgnoreCase(secondString)); //prints true

Locale locale = Locale.forLanguageTag("tr-TR");

                   secondString.toLowerCase(locale))); //prints false

Live demo

Do not use the == operator to compare Strings

Unless you can guarantee that all strings have been interned (see below), you should not use the == or != operators to compare Strings. These operators actually test references, and since multiple String objects can represent the same String, this is liable to give the wrong answer.

Instead, use the String.equals(Object) method, which will compare the String objects based on their values. For a detailed explanation, please refer to Pitfall: using == to compare strings.

Comparing Strings in a switch statement

Java SE 7

As of Java 1.7, it is possible to compare a String variable to literals in a switch statement. Make sure that the String is not null, otherwise it will always throw a NullPointerException. Values are compared using String.equals, i.e. case sensitive.

String stringToSwitch = "A";

switch (stringToSwitch) {
    case "a":
    case "A":
        System.out.println("A"); //the code goes here
    case "B":

Live demo

Comparing Strings with constant values

When comparing a String to a constant value, you can put the constant value on the left side of equals to ensure that you won't get a NullPointerException if the other String is null.


While foo.equals("baz") will throw a NullPointerException if foo is null, "baz".equals(foo) will evaluate to false.

Java SE 7

A more readable alternative is to use Objects.equals(), which does a null check on both parameters: Objects.equals(foo, "baz").

(Note: It is debatable as to whether it is better to avoid NullPointerExceptions in general, or let them happen and then fix the root cause; see here and here. Certainly, calling the avoidance strategy "best practice" is not justifiable.)

String orderings

The String class implements Comparable<String> with the String.compareTo method (as described at the start of this example). This makes the natural ordering of String objects case-sensitive order. The String class provide a Comparator<String> constant called CASE_INSENSITIVE_ORDER suitable for case-insensitive sorting.

Comparing with interned Strings

The Java Language Specification (JLS 3.10.6) states the following:

"Moreover, a string literal always refers to the same instance of class String. This is because string literals - or, more generally, strings that are the values of constant expressions - are interned so as to share unique instances, using the method String.intern."

This means it is safe to compare references to two string literals using ==. Moreover, the same is true for references to String objects that have been produced using the String.intern() method.

For example:

String strObj = new String("Hello!");
String str = "Hello!";

// The two string references point two strings that are equal
if (strObj.equals(str)) {
    System.out.println("The strings are equal");

// The two string references do not point to the same object
if (strObj != str) {
    System.out.println("The strings are not the same object");

// If we intern a string that is equal to a given literal, the result is
// a string that has the same reference as the literal.
String internedStr = strObj.intern();

if (internedStr == str) {
    System.out.println("The interned string and the literal are the same object");

Behind the scenes, the interning mechanism maintains a hash table that contains all interned strings that are still reachable. When you call intern() on a String, the method looks up the object in the hash table:

  • If the string is found, then that value is returned as the interned string.
  • Otherwise, a copy of the string is added to the hash table and that string is returned as the interned string.

It is possible to use interning to allow strings to be compared using ==. However, there are significant problems with doing this; see Pitfall - Interning strings so that you can use == is a bad idea for details. It is not recommended in most cases.

Changing the case of characters within a String

The String type provides two methods for converting strings between upper case and lower case:

These methods both return the converted strings as new String instances: the original String objects are not modified because String is immutable in Java. See this for more on immutability : Immutability of Strings in Java

String string = "This is a Random String";
String upper = string.toUpperCase();
String lower = string.toLowerCase();

System.out.println(string);   // prints "This is a Random String"
System.out.println(lower);    // prints "this is a random string"
System.out.println(upper);    // prints "THIS IS A RANDOM STRING"

Non-alphabetic characters, such as digits and punctuation marks, are unaffected by these methods. Note that these methods may also incorrectly deal with certain Unicode characters under certain conditions.

Note: These methods are locale-sensitive, and may produce unexpected results if used on strings that are intended to be interpreted independent of the locale. Examples are programming language identifiers, protocol keys, and HTML tags.

For instance, "TITLE".toLowerCase() in a Turkish locale returns "tıtle", where ı (\u0131) is the LATIN SMALL LETTER DOTLESS I character. To obtain correct results for locale insensitive strings, pass Locale.ROOT as a parameter to the corresponding case converting method (e.g. toLowerCase(Locale.ROOT) or toUpperCase(Locale.ROOT)).

Although using Locale.ENGLISH is also correct for most cases, the language invariant way is Locale.ROOT.

A detailed list of Unicode characters that require special casing can be found on the Unicode Consortium website.

Changing case of a specific character within an ASCII string:

To change the case of a specific character of an ASCII string following algorithm can be used:


  1. Declare a string.
  2. Input the string.
  3. Convert the string into a character array.
  4. Input the character that is to be searched.
  5. Search for the character into the character array.
  6. If found,check if the character is lowercase or uppercase.
    • If Uppercase, add 32 to the ASCII code of the character.
    • If Lowercase, subtract 32 from the ASCII code of the character.
  7. Change the original character from the Character array.
  8. Convert the character array back into the string.

Voila, the Case of the character is changed.

An example of the code for the algorithm is:

Scanner scanner = new Scanner(;
System.out.println("Enter the String");
String s =;
char[] a = s.toCharArray();
System.out.println("Enter the character you are looking for");
String c =;
char d = c.charAt(0);

for (int i = 0; i <= s.length(); i++) {
    if (a[i] == d) {
        if (d >= 'a' && d <= 'z') {
            d -= 32;
        } else if (d >= 'A' && d <= 'Z') {
            d += 32;
        a[i] = d;
s = String.valueOf(a);

Finding a String Within Another String

To check whether a particular String a is being contained in a String b or not, we can use the method String.contains() with the following syntax:

b.contains(a); // Return true if a is contained in b, false otherwise

The String.contains() method can be used to verify if a CharSequence can be found in the String. The method looks for the String a in the String b in a case-sensitive way.

String str1 = "Hello World";
String str2 = "Hello";
String str3 = "helLO";

System.out.println(str1.contains(str2)); //prints true
System.out.println(str1.contains(str3)); //prints false

Live Demo on Ideone

To find the exact position where a String starts within another String, use String.indexOf():

String s = "this is a long sentence";
int i = s.indexOf('i');    // the first 'i' in String is at index 2
int j = s.indexOf("long"); // the index of the first occurrence of "long" in s is 10
int k = s.indexOf('z');    // k is -1 because 'z' was not found in String s
int h = s.indexOf("LoNg"); // h is -1 because "LoNg" was not found in String s

Live Demo on Ideone

The String.indexOf() method returns the first index of a char or String in another String. The method returns -1 if it is not found.

Note: The String.indexOf() method is case sensitive.

Example of search ignoring the case:

String str1 = "Hello World";
String str2 = "wOr";
str1.indexOf(str2);                               // -1
str1.toLowerCase().contains(str2.toLowerCase());  // true
str1.toLowerCase().indexOf(str2.toLowerCase());   // 6

Live Demo on Ideone

Getting the length of a String

In order to get the length of a String object, call the length() method on it. The length is equal to the number of UTF-16 code units (chars) in the string.

String str = "Hello, World!";
System.out.println(str.length()); // Prints out 13

Live Demo on Ideone

A char in a String is UTF-16 value. Unicode codepoints whose values are ≥ 0x1000 (for example, most emojis) use two char positions. To count the number of Unicode codepoints in a String, regardless of whether each codepoint fits in a UTF-16 char value, you can use the codePointCount method:

int length = str.codePointCount(0, str.length());

You can also use a Stream of codepoints, as of Java 8:

int length = str.codePoints().count();


String s = "this is an example";
String a = s.substring(11); // a will hold the string starting at character 11 until the end ("example")
String b = s.substring(5, 10); // b will hold the string starting at character 5 and ending right before character 10 ("is an")
String b = s.substring(5, b.length()-3); // b will hold the string starting at character 5 ending right before b' s lenght is out of 3  ("is an exam")

Substrings may also be applied to slice and add/replace character into its original String. For instance, you faced a Chinese date containing Chinese characters but you want to store it as a well format Date String.

String datestring = "2015年11月17日"
datestring = datestring.substring(0, 4) + "-" + datestring.substring(5,7) + "-" + datestring.substring(8,10);
//Result will be 2015-11-17

The substring method extracts a piece of a String. When provided one parameter, the parameter is the start and the piece extends until the end of the String. When given two parameters, the first parameter is the starting character and the second parameter is the index of the character right after the end (the character at the index is not included). An easy way to check is the subtraction of the first parameter from the second should yield the expected length of the string.

Java SE 7

In JDK <7u6 versions the substring method instantiates a String that shares the same backing char[] as the original String and has the internal offset and count fields set to the result start and length. Such sharing may cause memory leaks, that can be prevented by calling new String(s.substring(...)) to force creation of a copy, after which the char[] can be garbage collected.

Java SE 7

From JDK 7u6 the substring method always copies the entire underlying char[] array, making the complexity linear compared to the previous constant one but guaranteeing the absence of memory leaks at the same time.

Getting the nth character in a String

String str = "My String";

System.out.println(str.charAt(0)); // "M"
System.out.println(str.charAt(1)); // "y"
System.out.println(str.charAt(2)); // " "
System.out.println(str.charAt(str.length-1)); // Last character "g"

To get the nth character in a string, simply call charAt(n) on a String, where n is the index of the character you would like to retrieve

NOTE: index n is starting at 0, so the first element is at n=0.

Platform independent new line separator

Since the new line separator varies from platform to platform (e.g. \n on Unix-like systems or \r\n on Windows) it is often necessary to have a platform-independent way of accessing it. In Java it can be retrieved from a system property:

Java SE 7

Because the new line separator is so commonly needed, from Java 7 on a shortcut method returning exactly the same result as the code above is available:


Note: Since it is very unlikely that the new line separator changes during the program's execution, it is a good idea to store it in in a static final variable instead of retrieving it from the system property every time it is needed.

When using String.format, use %n rather than \n or '\r\n' to output a platform independent new line separator.

System.out.println(String.format('line 1: %s.%nline 2: %s%n', lines[0],lines[1]));

Adding toString() method for custom objects

Suppose you have defined the following Person class:

public class Person {

    String name;
    int age;
    public Person (int age, String name) {
        this.age = age; = name;

If you instantiate a new Person object:

Person person = new Person(25, "John");

and later in your code you use the following statement in order to print the object:


Live Demo on Ideone

you'll get an output similar to the following:

[email protected]

This is the result of the implementation of the toString() method defined in the Object class, a superclass of Person. The documentation of Object.toString() states:

The toString method for class Object returns a string consisting of the name of the class of which the object is an instance, the at-sign character `@', and the unsigned hexadecimal representation of the hash code of the object. In other words, this method returns a string equal to the value of:

getClass().getName() + '@' + Integer.toHexString(hashCode())

So, for meaningful output, you'll have to override the toString() method:

public String toString() {
    return "My name is " + + " and my age is " + this.age;

Now the output will be:

My name is John and my age is 25

You can also write


Live Demo on Ideone

In fact, println() implicitly invokes the toString method on the object.

Splitting Strings

You can split a String on a particular delimiting character or a Regular Expression, you can use the String.split() method that has the following signature:

public String[] split(String regex)

Note that delimiting character or regular expression gets removed from the resulting String Array.

Example using delimiting character:

String lineFromCsvFile = "Mickey;Bolton;12345;121216";
String[] dataCells = lineFromCsvFile.split(";");
// Result is dataCells = { "Mickey", "Bolton", "12345", "121216"};

Example using regular expression:

String lineFromInput = "What    do you need    from me?";
String[] words = lineFromInput.split("\\s+"); // one or more space chars
// Result is words = {"What", "do", "you", "need", "from", "me?"};

You can even directly split a String literal:

String[] firstNames = "Mickey, Frank, Alicia, Tom".split(", ");
// Result is firstNames = {"Mickey", "Frank", "Alicia", "Tom"};

Warning: Do not forget that the parameter is always treated as a regular expression.

"aaa.bbb".split("."); // This returns an empty array

In the previous example . is treated as the regular expression wildcard that matches any character, and since every character is a delimiter, the result is an empty array.

Splitting based on a delimiter which is a regex meta-character

The following characters are considered special (aka meta-characters) in regex

  < > - = ! ( ) [ ] { } \ ^ $ | ? * + . 

To split a string based on one of the above delimiters, you need to either escape them using \\ or use Pattern.quote():

  • Using Pattern.quote():

     String s = "a|b|c";
     String regex = Pattern.quote("|");
     String[] arr = s.split(regex);
  • Escaping the special characters:

     String s = "a|b|c";
     String[] arr = s.split("\\|");

Split removes empty values

split(delimiter) by default removes trailing empty strings from result array. To turn this mechanism off we need to use overloaded version of split(delimiter, limit) with limit set to negative value like

String[] split = data.split("\\|", -1);

split(regex) internally returns result of split(regex, 0).

The limit parameter controls the number of times the pattern is applied and therefore affects the length of the resulting array.
If the limit n is greater than zero then the pattern will be applied at most n - 1 times, the array's length will be no greater than n, and the array's last entry will contain all input beyond the last matched delimiter.
If n is negative, then the pattern will be applied as many times as possible and the array can have any length.
If n is zero then the pattern will be applied as many times as possible, the array can have any length, and trailing empty strings will be discarded.

Splitting with a StringTokenizer

Besides the split() method Strings can also be split using a StringTokenizer.

StringTokenizer is even more restrictive than String.split(), and also a bit harder to use. It is essentially designed for pulling out tokens delimited by a fixed set of characters (given as a String). Each character will act as a separator. Because of this restriction, it's about twice as fast as String.split().

Default set of characters are empty spaces (\t\n\r\f). The following example will print out each word separately.

String str = "the lazy fox jumped over the brown fence";
StringTokenizer tokenizer = new StringTokenizer(str);
while (tokenizer.hasMoreTokens()) {

This will print out:


You can use different character sets for separation.

String str = "jumped over";
// In this case character `u` and `e` will be used as delimiters 
StringTokenizer tokenizer = new StringTokenizer(str, "ue");
while (tokenizer.hasMoreTokens()) {

This will print out:

d ov

Joining Strings with a delimiter

Java SE 8

An array of strings can be joined using the static method String.join():

String[] elements = { "foo", "bar", "foobar" };
String singleString = String.join(" + ", elements);

System.out.println(singleString); // Prints "foo + bar + foobar"     

Similarly, there's an overloaded String.join() method for Iterables.

To have a fine-grained control over joining, you may use StringJoiner class:

StringJoiner sj = new StringJoiner(", ", "[", "]"); 
    // The last two arguments are optional, 
    // they define prefix and suffix for the result string


System.out.println(sj); // Prints "[foo, bar, foobar]"

To join a stream of strings, you may use the joining collector:

Stream<String> stringStream = Stream.of("foo", "bar", "foobar");
String joined = stringStream.collect(Collectors.joining(", "));
System.out.println(joined); // Prints "foo, bar, foobar"

There's an option to define prefix and suffix here as well:

Stream<String> stringStream = Stream.of("foo", "bar", "foobar");
String joined = stringStream.collect(Collectors.joining(", ", "{", "}"));
System.out.println(joined); // Prints "{foo, bar, foobar}"

Reversing Strings

There are a couple ways you can reverse a string to make it backwards.

  1. StringBuilder/StringBuffer:

     String code = "code";
     StringBuilder sb = new StringBuilder(code); 
     code = sb.reverse().toString();
  2. Char array:

    String code = "code";
    char[] array = code.toCharArray();
    for (int index = 0, mirroredIndex = array.length - 1; index < mirroredIndex; index++, mirroredIndex--) {
        char temp = array[index];
        array[index] = array[mirroredIndex];
        array[mirroredIndex] = temp;
    // print reversed
    System.out.println(new String(array));

Counting occurrences of a substring or character in a string

countMatches method from org.apache.commons.lang3.StringUtils is typically used to count occurences of a substring or character in a String:

import org.apache.commons.lang3.StringUtils;

String text = "One fish, two fish, red fish, blue fish";

// count occurrences of a substring
String stringTarget = "fish";
int stringOccurrences = StringUtils.countMatches(text, stringTarget); // 4

// count occurrences of a char
char charTarget = ',';
int charOccurrences = StringUtils.countMatches(text, charTarget); // 3

Otherwise for does the same with standard Java API's you could use Regular Expressions:

import java.util.regex.Matcher;
import java.util.regex.Pattern;
String text = "One fish, two fish, red fish, blue fish";
System.out.println(countStringInString("fish", text)); // prints 4
System.out.println(countStringInString(",", text)); // prints 3

public static int countStringInString(String search, String text) {
    Pattern pattern = Pattern.compile(search);
    Matcher matcher = pattern.matcher(text);
    int stringOccurrences = 0;
    while (matcher.find()) {
    return stringOccurrences;

String concatenation and StringBuilders

String concatenation can be performed using the + operator. For example:

String s1 = "a";
String s2 = "b";
String s3 = "c";
String s = s1 + s2 + s3; // abc

Normally a compiler implementation will perform the above concatenation using methods involving a StringBuilder under the hood. When compiled, the code would look similar to the below:

StringBuilder sb = new StringBuilder("a");
String s = sb.append("b").append("c").toString();

StringBuilder has several overloaded methods for appending different types, for example, to append an int instead of a String. For example, an implementation can convert:

String s1 = "a";
String s2 = "b";    
String s = s1 + s2 + 2; // ab2

to the following:

StringBuilder sb = new StringBuilder("a");
String s = sb.append("b").append(2).toString();

The above examples illustrate a simple concatenation operation that is effectively done in a single place in the code. The concatenation involves a single instance of the StringBuilder. In some cases, a concatenation is carried out in a cumulative way such as in a loop:

String result = "";
for(int i = 0; i < array.length; i++) {
    result += extractElement(array[i]);
return result;

In such cases, the compiler optimization is usually not applied, and each iteration will create a new StringBuilder object. This can be optimized by explicitly transforming the code to use a single StringBuilder:

StringBuilder result = new StringBuilder();
for(int i = 0; i < array.length; i++) {
return result.toString();

A StringBuilder will be initialized with an empty space of only 16 characters. If you know in advance that you will be building larger strings, it can be beneficial to initialize it with sufficient size in advance, so that the internal buffer does not need to be resized:

StringBuilder buf = new StringBuilder(30); // Default is 16 characters
buf.append("0123456789"); // Would cause a reallocation of the internal buffer otherwise
String result = buf.toString(); // Produces a 20-chars copy of the string

If you are producing many strings, it is advisable to reuse StringBuilders:

StringBuilder buf = new StringBuilder(100);
for (int i = 0; i < 100; i++) {
    buf.setLength(0); // Empty buffer
    buf.append("This is line ").append(i).append('\n');

If (and only if) multiple threads are writing to the same buffer, use StringBuffer, which is a synchronized version of StringBuilder. But because usually only a single thread writes to a buffer, it is usually faster to use StringBuilder without synchronization.

Using concat() method:

String string1 = "Hello ";
String string2 = "world";
String string3 = string1.concat(string2);  // "Hello world"

This returns a new string that is string1 with string2 added to it at the end. You can also use the concat() method with string literals, as in:

"My name is ".concat("Buyya");

Replacing parts of Strings

Two ways to replace: by regex or by exact match.

Note: the original String object will be unchanged, the return value holds the changed String.

Exact match

Replace single character with another single character:

String replace(char oldChar, char newChar) 

Returns a new string resulting from replacing all occurrences of oldChar in this string with newChar.

String s = "popcorn";



Replace sequence of characters with another sequence of characters:

String replace(CharSequence target, CharSequence replacement) 

Replaces each substring of this string that matches the literal target sequence with the specified literal replacement sequence.

String s = "metal petal et al.";


metallica petallica et al.


Note: the grouping uses the $ character to reference the groups, like $1.

Replace all matches:

String replaceAll(String regex, String replacement) 

Replaces each substring of this string that matches the given regular expression with the given replacement.

String s = "spiral metal petal et al.";


spiral metallica petallica et al.

Replace first match only:

String replaceFirst(String regex, String replacement) 

Replaces the first substring of this string that matches the given regular expression with the given replacement

String s = "spiral metal petal et al.";


spiral metallica petal et al.

Remove Whitespace from the Beginning and End of a String

The trim() method returns a new String with the leading and trailing whitespace removed.

String s = new String("   Hello World!!  ");
String t = s.trim();  // t = "Hello World!!"

If you trim a String that doesn't have any whitespace to remove, you will be returned the same String instance.

Note that the trim() method has its own notion of whitespace, which differs from the notion used by the Character.isWhitespace() method:

  • All ASCII control characters with codes U+0000 to U+0020 are considered whitespace and are removed by trim(). This includes U+0020 'SPACE', U+0009 'CHARACTER TABULATION', U+000A 'LINE FEED' and U+000D 'CARRIAGE RETURN' characters, but also the characters like U+0007 'BELL'.

  • Unicode whitespace like U+00A0 'NO-BREAK SPACE' or U+2003 'EM SPACE' are not recognized by trim().

String pool and heap storage

Like many Java objects, all String instances are created on the heap, even literals. When the JVM finds a String literal that has no equivalent reference in the heap, the JVM creates a corresponding String instance on the heap and it also stores a reference to the newly created String instance in the String pool. Any other references to the same String literal are replaced with the previously created String instance in the heap.

Let's look at the following example:

class Strings
    public static void main (String[] args)
        String a = "alpha";
        String b = "alpha";
        String c = new String("alpha");

        //All three strings are equivalent
        System.out.println(a.equals(b) && b.equals(c));

        //Although only a and b reference the same heap object
        System.out.println(a == b);
        System.out.println(a != c);
        System.out.println(b != c);

The output of the above is:


Diagram of Java Heap and String pool When we use double quotes to create a String, it first looks for String with 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 other String object from string pool having same value.

The String pool itself is also created on the heap.

Java SE 7

Before Java 7, String literals were stored in the runtime constant pool in the method area of PermGen, that had a fixed size.

The String pool also resided in PermGen.

Java SE 7

RFC: 6962931

In JDK 7, interned strings are no longer allocated in the permanent generation of the Java heap, but are instead allocated in the main part of the Java heap (known as the young and old generations), along with the other objects created by the application. This change will result in more data residing in the main Java heap, and less data in the permanent generation, and thus may require heap sizes to be adjusted. Most applications will see only relatively small differences in heap usage due to this change, but larger applications that load many classes or make heavy use of the String.intern() method will see more significant differences.

Case insensitive switch

Java SE 7

switch itself can not be parameterised to be case insensitive, but if absolutely required, can behave insensitive to the input string by using toLowerCase() or toUpperCase:

switch (myString.toLowerCase()) {
     case "case1" :
     case "case2" :


  • Locale might affect how changing cases happen!
  • Care must be taken not to have any uppercase characters in the labels - those will never get executed!

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