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This question already has answers here:
How do I compare strings in Java?
(23 answers)
Closed 9 years ago.
This code separates a string into tokens and stores them in an array of strings, and then compares a variable with the first home ... why isn't it working?
public static void main(String...aArguments) throws IOException {
String usuario = "Jorman";
String password = "14988611";
String strDatos = "Jorman 14988611";
StringTokenizer tokens = new StringTokenizer(strDatos, " ");
int nDatos = tokens.countTokens();
String[] datos = new String[nDatos];
int i = 0;
while (tokens.hasMoreTokens()) {
String str = tokens.nextToken();
datos[i] = str;
i++;
}
//System.out.println (usuario);
if ((datos[0] == usuario)) {
System.out.println("WORKING");
}
}
Use the string.equals(Object other) function to compare strings, not the == operator.
The function checks the actual contents of the string, the == operator checks whether the references to the objects are equal. Note that string constants are usually "interned" such that two constants with the same value can actually be compared with ==, but it's better not to rely on that.
if (usuario.equals(datos[0])) {
...
}
NB: the compare is done on 'usuario' because that's guaranteed non-null in your code, although you should still check that you've actually got some tokens in the datos array otherwise you'll get an array-out-of-bounds exception.
Meet Jorman
Jorman is a successful businessman and has 2 houses.
But others don't know that.
Is it the same Jorman?
When you ask neighbours from either Madison or Burke streets, this is the only thing they can say:
Using the residence alone, it's tough to confirm that it's the same Jorman. Since they're 2 different addresses, it's just natural to assume that those are 2 different persons.
That's how the operator == behaves. So it will say that datos[0]==usuario is false, because it only compares the addresses.
An Investigator to the Rescue
What if we sent an investigator? We know that it's the same Jorman, but we need to prove it. Our detective will look closely at all physical aspects. With thorough inquiry, the agent will be able to conclude whether it's the same person or not. Let's see it happen in Java terms.
Here's the source code of String's equals() method:
It compares the Strings character by character, in order to come to a conclusion that they are indeed equal.
That's how the String equals method behaves. So datos[0].equals(usuario) will return true, because it performs a logical comparison.
It's good to notice that in some cases use of "==" operator can lead to the expected result, because the way how java handles strings - string literals are interned (see String.intern()) during compilation - so when you write for example "hello world" in two classes and compare those strings with "==" you could get result: true, which is expected according to specification; when you compare same strings (if they have same value) when the first one is string literal (ie. defined through "i am string literal") and second is constructed during runtime ie. with "new" keyword like new String("i am string literal"), the == (equality) operator returns false, because both of them are different instances of the String class.
Only right way is using .equals() -> datos[0].equals(usuario). == says only if two objects are the same instance of object (ie. have same memory address)
Update: 01.04.2013 I updated this post due comments below which are somehow right. Originally I declared that interning (String.intern) is side effect of JVM optimization. Although it certainly save memory resources (which was what i meant by "optimization") it is mainly feature of language
The == operator checks if the two references point to the same object or not.
.equals() checks for the actual string content (value).
Note that the .equals() method belongs to class Object (super class of all classes). You need to override it as per you class requirement, but for String it is already implemented and it checks whether two strings have the same value or not.
Case1)
String s1 = "Stack Overflow";
String s2 = "Stack Overflow";
s1 == s1; // true
s1.equals(s2); // true
Reason: String literals created without null are stored in the string pool in the permgen area of the heap. So both s1 and s2 point to the same object in the pool.
Case2)
String s1 = new String("Stack Overflow");
String s2 = new String("Stack Overflow");
s1 == s2; // false
s1.equals(s2); // true
Reason: If you create a String object using the `new` keyword a separate space is allocated to it on the heap.
equals() function is a method of Object class which should be overridden by programmer. String class overrides it to check if two strings are equal i.e. in content and not reference.
== operator checks if the references of both the objects are the same.
Consider the programs
String abc = "Awesome" ;
String xyz = abc;
if(abc == xyz)
System.out.println("Refers to same string");
Here the abc and xyz, both refer to same String "Awesome". Hence the expression (abc == xyz) is true.
String abc = "Hello World";
String xyz = "Hello World";
if(abc == xyz)
System.out.println("Refers to same string");
else
System.out.println("Refers to different strings");
if(abc.equals(xyz))
System.out.prinln("Contents of both strings are same");
else
System.out.prinln("Contents of strings are different");
Here abc and xyz are two different strings with the same content "Hello World". Hence here the expression (abc == xyz) is false where as (abc.equals(xyz)) is true.
Hope you understood the difference between == and <Object>.equals()
Thanks.
== tests for reference equality.
.equals() tests for value equality.
Consequently, if you actually want to test whether two strings have the same value you should use .equals() (except in a few situations where you can guarantee that two strings with the same value will be represented by the same object eg: String interning).
== is for testing whether two strings are the same Object.
// These two have the same value
new String("test").equals("test") ==> true
// ... but they are not the same object
new String("test") == "test" ==> false
// ... neither are these
new String("test") == new String("test") ==> false
// ... but these are because literals are interned by
// the compiler and thus refer to the same object
"test" == "test" ==> true
// concatenation of string literals happens at compile time resulting in same objects
"test" == "te" + "st" ==> true
// but .substring() is invoked at runtime, generating distinct objects
"test" == "!test".substring(1) ==> false
It is important to note that == is much cheaper than equals() (a single pointer comparision instead of a loop), thus, in situations where it is applicable (i.e. you can guarantee that you are only dealing with interned strings) it can present an important performance improvement. However, these situations are rare.
Instead of
datos[0] == usuario
use
datos[0].equals(usuario)
== compares the reference of the variable where .equals() compares the values which is what you want.
Let's analyze the following Java, to understand the identity and equality of Strings:
public static void testEquality(){
String str1 = "Hello world.";
String str2 = "Hello world.";
if (str1 == str2)
System.out.print("str1 == str2\n");
else
System.out.print("str1 != str2\n");
if(str1.equals(str2))
System.out.print("str1 equals to str2\n");
else
System.out.print("str1 doesn't equal to str2\n");
String str3 = new String("Hello world.");
String str4 = new String("Hello world.");
if (str3 == str4)
System.out.print("str3 == str4\n");
else
System.out.print("str3 != str4\n");
if(str3.equals(str4))
System.out.print("str3 equals to str4\n");
else
System.out.print("str3 doesn't equal to str4\n");
}
When the first line of code String str1 = "Hello world." executes, a string \Hello world."
is created, and the variable str1 refers to it. Another string "Hello world." will not be created again when the next line of code executes because of optimization. The variable str2 also refers to the existing ""Hello world.".
The operator == checks identity of two objects (whether two variables refer to same object). Since str1 and str2 refer to same string in memory, they are identical to each other. The method equals checks equality of two objects (whether two objects have same content). Of course, the content of str1 and str2 are same.
When code String str3 = new String("Hello world.") executes, a new instance of string with content "Hello world." is created, and it is referred to by the variable str3. And then another instance of string with content "Hello world." is created again, and referred to by
str4. Since str3 and str4 refer to two different instances, they are not identical, but their
content are same.
Therefore, the output contains four lines:
Str1 == str2
Str1 equals str2
Str3! = str4
Str3 equals str4
You should use string equals to compare two strings for equality, not operator == which just compares the references.
It will also work if you call intern() on the string before inserting it into the array.
Interned strings are reference-equal (==) if and only if they are value-equal (equals().)
public static void main (String... aArguments) throws IOException {
String usuario = "Jorman";
String password = "14988611";
String strDatos="Jorman 14988611";
StringTokenizer tokens=new StringTokenizer(strDatos, " ");
int nDatos=tokens.countTokens();
String[] datos=new String[nDatos];
int i=0;
while(tokens.hasMoreTokens()) {
String str=tokens.nextToken();
datos[i]= str.intern();
i++;
}
//System.out.println (usuario);
if(datos[0]==usuario) {
System.out.println ("WORKING");
}
Generally .equals is used for Object comparison, where you want to verify if two Objects have an identical value.
== for reference comparison (are the two Objects the same Object on the heap) & to check if the Object is null. It is also used to compare the values of primitive types.
== operator compares the reference of an object in Java. You can use string's equals method .
String s = "Test";
if(s.equals("Test"))
{
System.out.println("Equal");
}
If you are going to compare any assigned value of the string i.e. primitive string, both "==" and .equals will work, but for the new string object you should use only .equals, and here "==" will not work.
Example:
String a = "name";
String b = "name";
if(a == b) and (a.equals(b)) will return true.
But
String a = new String("a");
In this case if(a == b) will return false
So it's better to use the .equals operator...
The == operator is a simple comparison of values.
For object references the (values) are the (references). So x == y returns true if x and y reference the same object.
I know this is an old question but here's how I look at it (I find very useful):
Technical explanations
In Java, all variables are either primitive types or references.
(If you need to know what a reference is: "Object variables" are just pointers to objects. So with Object something = ..., something is really an address in memory (a number).)
== compares the exact values. So it compares if the primitive values are the same, or if the references (addresses) are the same. That's why == often doesn't work on Strings; Strings are objects, and doing == on two string variables just compares if the address is same in memory, as others have pointed out. .equals() calls the comparison method of objects, which will compare the actual objects pointed by the references. In the case of Strings, it compares each character to see if they're equal.
The interesting part:
So why does == sometimes return true for Strings? Note that Strings are immutable. In your code, if you do
String foo = "hi";
String bar = "hi";
Since strings are immutable (when you call .trim() or something, it produces a new string, not modifying the original object pointed to in memory), you don't really need two different String("hi") objects. If the compiler is smart, the bytecode will read to only generate one String("hi") object. So if you do
if (foo == bar) ...
right after, they're pointing to the same object, and will return true. But you rarely intend this. Instead, you're asking for user input, which is creating new strings at different parts of memory, etc. etc.
Note: If you do something like baz = new String(bar) the compiler may still figure out they're the same thing. But the main point is when the compiler sees literal strings, it can easily optimize same strings.
I don't know how it works in runtime, but I assume the JVM doesn't keep a list of "live strings" and check if a same string exists. (eg if you read a line of input twice, and the user enters the same input twice, it won't check if the second input string is the same as the first, and point them to the same memory). It'd save a bit of heap memory, but it's so negligible the overhead isn't worth it. Again, the point is it's easy for the compiler to optimize literal strings.
There you have it... a gritty explanation for == vs. .equals() and why it seems random.
#Melkhiah66 You can use equals method instead of '==' method to check the equality.
If you use intern() then it checks whether the object is in pool if present then returns
equal else unequal. equals method internally uses hashcode and gets you the required result.
public class Demo
{
public static void main(String[] args)
{
String str1 = "Jorman 14988611";
String str2 = new StringBuffer("Jorman").append(" 14988611").toString();
String str3 = str2.intern();
System.out.println("str1 == str2 " + (str1 == str2)); //gives false
System.out.println("str1 == str3 " + (str1 == str3)); //gives true
System.out.println("str1 equals str2 " + (str1.equals(str2))); //gives true
System.out.println("str1 equals str3 " + (str1.equals(str3))); //gives true
}
}
The .equals() will check if the two strings have the same value and return the boolean value where as the == operator checks to see if the two strings are the same object.
Someone said on a post higher up that == is used for int and for checking nulls.
It may also be used to check for Boolean operations and char types.
Be very careful though and double check that you are using a char and not a String.
for example
String strType = "a";
char charType = 'a';
for strings you would then check
This would be correct
if(strType.equals("a")
do something
but
if(charType.equals('a')
do something else
would be incorrect, you would need to do the following
if(charType == 'a')
do something else
a==b
Compares references, not values. The use of == with object references is generally limited to the following:
Comparing to see if a reference is null.
Comparing two enum values. This works because there is only one object for each enum constant.
You want to know if two references are to the same object
"a".equals("b")
Compares values for equality. Because this method is defined in the Object class, from which all other classes are derived, it's automatically defined for every class. However, it doesn't perform an intelligent comparison for most classes unless the class overrides it. It has been defined in a meaningful way for most Java core classes. If it's not defined for a (user) class, it behaves the same as ==.
Use Split rather than tokenizer,it will surely provide u exact output
for E.g:
string name="Harry";
string salary="25000";
string namsal="Harry 25000";
string[] s=namsal.split(" ");
for(int i=0;i<s.length;i++)
{
System.out.println(s[i]);
}
if(s[0].equals("Harry"))
{
System.out.println("Task Complete");
}
After this I am sure you will get better results.....
This question already has answers here:
How do I compare strings in Java?
(23 answers)
What is the difference between "text" and new String("text")?
(13 answers)
Closed 3 years ago.
public class Test {
public static void main(String[] args)
{
String s1 = "HELLO";
String s2 = "HELLO";
System.out.println(s1 == s2); // true
}
}
But when I use :
public class Test {
public static void main(String[] args)
{
String s1 = new String("HELLO");
String s2 = new String("HELLO");
System.out.println(s1 == s2); // false
}
}
Can anybody please explain the difference here? Thankyou!
In the first example
String s1 = "HELLO";
String s2 = "HELLO";
the values of s1 and s2 are compile-time constants. Thus, the compiler does only generate a single String-object, holding the value "HELLO" and assings it to both s1 and s2. This is a special case of Common Subexpression Elimination, a well-known compiler optimization. Thus s1 == s2 returns true.
In the second example, two different Strings are constructed explicitly through new. Thus, they have to be separate objects per the semantics of new.
I created an Ideone demo a while back that highlights some cases that show this behaviour.
You can enforce that the same String is return by using String::intern():
String s1 = new String("HELLO").intern();
String s2 = new String("HELLO").intern();
System.out.println(s1 == s2); // will print "true";
Ideone demo
In case of String literal,before creating new Object in String Constant Pool ,JVM will check already same Object persist in SCP area or not if yes it will point to same object instead of creating new Object.Hence, below code s1 == s2 is true
String s1 = "HELLO";
String s2 = "HELLO";
System.out.println(s1 == s2); // true
but we are creating new object by using new keyword, it will create object in heap area, hence s1 and s2 are pointing to two different object, hence it is return false
== tests for reference equality (whether they are the same object).
.equals() tests for value equality (whether they are logically "equal").
from here How do I compare strings in Java?
== compares the objects reference pointer. When 2 objects are same exact object it will be true.
Instantiating a string using double quotes uses the string pool, creates a string once and reuses it.
Instantiating a string wit new always creates a brand new string.
== tests for reference equality (whether they are the same object).
First Case
System.out.println(s1 == s2); // true
Because you are comparing literals that are interned by the compiler and thus refer to the same object. 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.
Second Case
System.out.println(s1 == s2); // false
You are comparing the Object reference which is different so you are getting false.
Please check https://docs.oracle.com/javase/specs/jls/se8/html/jls-15.html#jls-15.28
i first case u comparing two strings with its ASCII values. thats why...//true
and in second case you are comparing two functions/methods. thats why... //false
the first one is true because s1 and s2 refer to the same string literal in the method area, the memory references are the same. ( == checks just the references in string). when the same string literal is created more than once, only one copy of each distinct string value is stored.
the second one is false because s1 and s2 refer to two different objects in the heap. different objects always have different memory references.
According to String#intern(), intern method is supposed to return the String from the String pool if the String is found in String pool, otherwise a new string object will be added in String pool and the reference of this String is returned.
So i tried this:
String s1 = "Rakesh";
String s2 = "Rakesh";
String s3 = "Rakesh".intern();
if ( s1 == s2 ){
System.out.println("s1 and s2 are same"); // 1.
}
if ( s1 == s3 ){
System.out.println("s1 and s3 are same" ); // 2.
}
I was expecting that s1 and s3 are same will be printed as s3 is interned, and s1 and s2 are same will not be printed. But the result is: both lines are printed. So that means, by default String constants are interned. But if it is so, then why do we need the intern method? In other words when should we use this method?
Java automatically interns String literals. This means that in many cases, the == operator appears to work for Strings in the same way that it does for ints or other primitive values.
Since interning is automatic for String literals, the intern() method is to be used on Strings constructed with new String()
Using your example:
String s1 = "Rakesh";
String s2 = "Rakesh";
String s3 = "Rakesh".intern();
String s4 = new String("Rakesh");
String s5 = new String("Rakesh").intern();
if ( s1 == s2 ){
System.out.println("s1 and s2 are same"); // 1.
}
if ( s1 == s3 ){
System.out.println("s1 and s3 are same" ); // 2.
}
if ( s1 == s4 ){
System.out.println("s1 and s4 are same" ); // 3.
}
if ( s1 == s5 ){
System.out.println("s1 and s5 are same" ); // 4.
}
will return:
s1 and s2 are same
s1 and s3 are same
s1 and s5 are same
In all the cases besides of s4 variable, a value for which was explicitly created using new operator and where intern method was not used on it's result, it is a single immutable instance that's being returned JVM's string constant pool.
Refer to JavaTechniques "String Equality and Interning" for more information.
On a recent project, some huge data structures were set up with data that was read in from a database (and hence not String constants/literals) but with a huge amount of duplication. It was a banking application, and things like the names of a modest set (maybe 100 or 200) corporations appeared all over the place. The data structures were already large, and if all those corp names had been unique objects they would have overflowed memory. Instead, all the data structures had references to the same 100 or 200 String objects, thus saving lots of space.
Another small advantage of interned Strings is that == can be used (successfully!) to compare Strings if all involved strings are guaranteed to be interned. Apart from the leaner syntax, this is also a performance enhancement. But as others have pointed out, doing this harbors a great risk of introducing programming errors, so this should be done only as a desparate measure of last resort.
The downside is that interning a String takes more time than simply throwing it on the heap, and that the space for interned Strings may be limited, depending on the Java implementation. It's best done when you're dealing with a known reasonable number of Strings with many duplications.
I want to add my 2 cents on using == with interned strings.
The first thing String.equals does is this==object.
So although there is some miniscule performance gain ( you are not calling a method), from the maintainer point of view using == is a nightmare, because some interned strings have a tendency to become non-interned.
So I suggest not to rely on special case of == for interned strings, but always use equals as Gosling intended.
EDIT: interned becoming non-interned:
V1.0
public class MyClass
{
private String reference_val;
...
private boolean hasReferenceVal ( final String[] strings )
{
for ( String s : strings )
{
if ( s == reference_val )
{
return true;
}
}
return false;
}
private void makeCall ( )
{
final String[] interned_strings = { ... init with interned values ... };
if ( hasReference( interned_strings ) )
{
...
}
}
}
In version 2.0 maintainer decided to make hasReferenceVal public, without going into much detail that it expects an array of interned strings.
V2.0
public class MyClass
{
private String reference_val;
...
public boolean hasReferenceVal ( final String[] strings )
{
for ( String s : strings )
{
if ( s == reference_val )
{
return true;
}
}
return false;
}
private void makeCall ( )
{
final String[] interned_strings = { ... init with interned values ... };
if ( hasReference( interned_strings ) )
{
...
}
}
}
Now you have a bug, that may be very hard to find, because in majority of cases array contains literal values, and sometimes a non-literal string is used. If equals were used instead of == then hasReferenceVal would have still continue to work. Once again, performance gain is miniscule, but maintenance cost is high.
Learn Java String Intern - once for all
Strings in java are immutable objects by design. Therefore, two string objects even with same value will be different objects by default. However, if we wish to save memory, we could indicate to use same memory by a concept called string intern. The below rules would help you understand the concept in clear terms:
String class maintains an intern-pool which is initially empty. This pool must guarantee to contain string objects with only unique values.
All string literals having same value must be considered same memory-location object because they have otherwise no notion of distinction. Therefore, all such literals with same value will make a single entry in the intern-pool and will refer to same memory location.
Concatenation of two or more literals is also a literal. (Therefore rule #2 will be applicable for them)
Each string created as object (i.e. by any other method except as literal) will have different memory locations and will not make any entry in the intern-pool
Concatenation of literals with non-literals will make a non-literal. Thus, the resultant object will have a new memory location and will NOT make an entry in the intern-pool.
Invoking intern method on a string object, either creates a new object that enters the intern-pool or return an existing object from the pool that has same value. The invocation on any object which is not in the intern-pool, does NOT move the object to the pool. It rather creates another object that enters the pool.
Example:
String s1=new String ("abc");
String s2=new String ("abc");
If (s1==s2) //would return false by rule #4
If ("abc" == "a"+"bc" ) //would return true by rules #2 and #3
If ("abc" == s1 ) //would return false by rules #1,2 and #4
If ("abc" == s1.intern() ) //would return true by rules #1,2,4 and #6
If ( s1 == s2.intern() ) //wound return false by rules #1,4, and #6
Note: The motivational cases for string intern are not discussed here. However, saving of memory will definitely be one of the primary objectives.
String literals and constants are interned by default.
That is, "foo" == "foo" (declared by the String literals), but new String("foo") != new String("foo").
you should make out two period time which are compile time and runtime time.for example:
//example 1
"test" == "test" // --> true
"test" == "te" + "st" // --> true
//example 2
"test" == "!test".substring(1) // --> false
"test" == "!test".substring(1).intern() // --> true
in the one hand,in the example 1,we find the results are all return true,because in the compile time,the jvm will put the "test" to the pool of literal strings,if the jvm find "test" exists,then it will use the exists one,in example 1,the "test" strings are all point to the same memory address,so the example 1 will return true.
in the other hand,in the example 2,the method of substring() execute in the runtime time,
in the case of "test" == "!test".substring(1),the pool will create two string object,"test"
and "!test",so they are different reference objects,so this case will return false,in the case of "test" == "!test".substring(1).intern(),the method of intern() will put the ""!test".substring(1)" to the pool of literal strings,so in this case,they are same reference objects,so will return true.
http://en.wikipedia.org/wiki/String_interning
string interning is a method of storing only one copy of each distinct string value, which must be immutable. Interning strings makes some string processing tasks more time- or space-efficient at the cost of requiring more time when the string is created or interned. The distinct values are stored in a string intern pool.
Interned Strings avoid duplicate Strings. Interning saves RAM at the expense of more CPU time to detect and replace duplicate Strings. There is only one copy of each String that has been interned, no matter how many references point to it. Since Strings are immutable, if two different methods incidentally use the same String, they can share a copy of the same String. The process of converting duplicated Strings to shared ones is called interning.String.intern() gives you the address of the canonical master String. You can compare interned Strings with simple == (which compares pointers) instead of equals which compares the characters of the String one by one. Because Strings are immutable, the intern process is free to further save space, for example, by not creating a separate String literal for "pot" when it exists as a substring of some other literal such as "hippopotamus".
To see more http://mindprod.com/jgloss/interned.html
String s1 = "Anish";
String s2 = "Anish";
String s3 = new String("Anish");
/*
* When the intern method is invoked, if the pool already contains a
* string equal to this String object as determined by the
* method, then the string from the pool is
* returned. Otherwise, this String object is added to the
* pool and a reference to this String object is returned.
*/
String s4 = new String("Anish").intern();
if (s1 == s2) {
System.out.println("s1 and s2 are same");
}
if (s1 == s3) {
System.out.println("s1 and s3 are same");
}
if (s1 == s4) {
System.out.println("s1 and s4 are same");
}
OUTPUT
s1 and s2 are same
s1 and s4 are same
String p1 = "example";
String p2 = "example";
String p3 = "example".intern();
String p4 = p2.intern();
String p5 = new String(p3);
String p6 = new String("example");
String p7 = p6.intern();
if (p1 == p2)
System.out.println("p1 and p2 are the same");
if (p1 == p3)
System.out.println("p1 and p3 are the same");
if (p1 == p4)
System.out.println("p1 and p4 are the same");
if (p1 == p5)
System.out.println("p1 and p5 are the same");
if (p1 == p6)
System.out.println("p1 and p6 are the same");
if (p1 == p6.intern())
System.out.println("p1 and p6 are the same when intern is used");
if (p1 == p7)
System.out.println("p1 and p7 are the same");
When two strings are created independently, intern() allows you to compare them and also it helps you in creating a reference in the string pool if the reference didn't exist before.
When you use String s = new String(hi), java creates a new instance of the string, but when you use String s = "hi", java checks if there is an instance of word "hi" in the code or not and if it exists, it just returns the reference.
Since comparing strings is based on reference, intern() helps in you creating a reference and allows you to compare the contents of the strings.
When you use intern() in the code, it clears of the space used by the string referring to the same object and just returns the reference of the already existing same object in memory.
But in case of p5 when you are using:
String p5 = new String(p3);
Only contents of p3 are copied and p5 is created newly. So it is not interned.
So the output will be:
p1 and p2 are the same
p1 and p3 are the same
p1 and p4 are the same
p1 and p6 are the same when intern is used
p1 and p7 are the same
public static void main(String[] args) {
// TODO Auto-generated method stub
String s1 = "test";
String s2 = new String("test");
System.out.println(s1==s2); //false
System.out.println(s1==s2.intern()); //true --> because this time compiler is checking from string constant pool.
}
string intern() method is used to create an exact copy of heap string object in string constant pool. The string objects in the string constant pool are automatically interned but string objects in heap are not. The main use of creating interns is to save the memory space and to perform faster comparison of string objects.
Source : What is string intern in java?
As you said, that string intern() method will first find from the String pool, if it finds, then it will return the object that points to that, or will add a new String into the pool.
String s1 = "Hello";
String s2 = "Hello";
String s3 = "Hello".intern();
String s4 = new String("Hello");
System.out.println(s1 == s2);//true
System.out.println(s1 == s3);//true
System.out.println(s1 == s4.intern());//true
The s1 and s2 are two objects pointing to the String pool "Hello", and using "Hello".intern() will find that s1 and s2. So "s1 == s3" returns true, as well as to the s3.intern().
By using heap object reference if we want to get corresponding string constant pool object reference, then we should go for intern()
String s1 = new String("Rakesh");
String s2 = s1.intern();
String s3 = "Rakesh";
System.out.println(s1 == s2); // false
System.out.println(s2 == s3); // true
Pictorial View
Step 1:
Object with data 'Rakesh' get created in heap and string constant pool. Also s1 is always pointing to heap object.
Step 2:
By using heap object reference s1, we are trying to get corresponding string constant pool object referenc s2, using intern()
Step 3:
Intentionally creating a object with data 'Rakesh' in string constant pool, referenced by name s3
As "==" operator meant for reference comparison.
Getting false for s1==s2
Getting true for s2==s3
Hope this help!!
The code below should not print "Bye", since == operator is used to compare references, but oddly enough, "Bye" is still printed. Why does this happen? I'm using Netbeans 6.9.1 as the IDE.
public class Test {
public static void main(String [] args) {
String test ="Hi";
if(test=="Hi"){
System.out.println("Bye");
}
}
}
This behavior is because of interning. The behavior is described in the docs for String#intern (including why it's showing up in your code even though you never call String#intern):
A pool of strings, initially empty, is maintained privately by the class String.
When the intern method is invoked, if the pool already contains a string equal to this String object as determined by the equals(Object) method, then the string from the pool is returned. Otherwise, this String object is added to the pool and a reference to this String object is returned.
It follows that for any two strings s and t, s.intern() == t.intern() is true if and only if s.equals(t) is true.
All literal strings and string-valued constant expressions are interned. String literals are defined in ยง3.10.5 of the Java Language Specification.
So for example:
public class Test {
private String s1 = "Hi";
public static void main(String [] args) {
new Test().test();
System.exit(0);
}
public void test() {
String s2 ="Hi";
String s3;
System.out.println("[statics] s2 == s1? " + (s2 == s1));
s3 = "H" + part2();
System.out.println("[before interning] s3 == s1? " + (s3 == s1));
s3 = s3.intern();
System.out.println("[after interning] s3 == s1? " + (s3 == s1));
System.exit(0);
}
protected String part2() {
return "i";
}
}
Output:
[statics] s2 == s1? true
[before interning] s3 == s1? false
[after interning] s3 == s1? true
Walking through that:
The literal assigned to s1 is automatically interned, so s1 ends up referring to a string in the pool.
The literal assigned to s2 is also auto-interned, and so s2 ends up pointing to the same instance s1 points to. This is fine even though the two bits of code may be completely unknown to each other, because Java's String instances are immutable. You can't change them. You can use methods like toLowerCase to get back a new string with changes, but the original you called toLowerCase (etc.) on remains unchanged. So they can safely be shared amongst unrelated code.
We create a new String instance via a runtime operation. Even though the new instance has the same sequence of characters as the interned one, it's a separate instance. The runtime doesn't intern dynamically-created strings automatically, because there's a cost involved: The work of finding the string in the pool. (Whereas when compiling, the compiler can take that cost onto itself.) So now we have two instances, the one s1 and s2 point to, and the one s3 points to. So the code shows that s3 != s1.
Then we explicitly intern s3. Perhaps it's a large string we're planning to hold onto for a long time, and we think it's likely that it's going to be duplicated in other places. So we accept the work of interning it in return for the potential memory savings. Since interning by definition means we may get back a new reference, we assign the result back to s3.
And we can see that indeed, s3 now points to the same instance s1 and s2 point to.
Hard-coded Strings are compiled into the JVM's String Table, which holds unique Strings - that is the compiler stores only one copy of "Hi", so you are comparing that same object, so == works.
If you actually create a new String using the constructor, like new String("Hi"), you will get a different object.
There is a String cache in java. Where as in this case the same object is returned from the cache which hold the same reference.
The main reason for that is because the "Hi" is picked up from String Pool. The immutable object must have some sort of cache so it can perform better. So String class is immutable and it uses String Pool for basic cache.
In this case, "Hi" is in the String pool and all the String having values of "Hi" will have the same reference for String Pool.
According to String#intern(), intern method is supposed to return the String from the String pool if the String is found in String pool, otherwise a new string object will be added in String pool and the reference of this String is returned.
So i tried this:
String s1 = "Rakesh";
String s2 = "Rakesh";
String s3 = "Rakesh".intern();
if ( s1 == s2 ){
System.out.println("s1 and s2 are same"); // 1.
}
if ( s1 == s3 ){
System.out.println("s1 and s3 are same" ); // 2.
}
I was expecting that s1 and s3 are same will be printed as s3 is interned, and s1 and s2 are same will not be printed. But the result is: both lines are printed. So that means, by default String constants are interned. But if it is so, then why do we need the intern method? In other words when should we use this method?
Java automatically interns String literals. This means that in many cases, the == operator appears to work for Strings in the same way that it does for ints or other primitive values.
Since interning is automatic for String literals, the intern() method is to be used on Strings constructed with new String()
Using your example:
String s1 = "Rakesh";
String s2 = "Rakesh";
String s3 = "Rakesh".intern();
String s4 = new String("Rakesh");
String s5 = new String("Rakesh").intern();
if ( s1 == s2 ){
System.out.println("s1 and s2 are same"); // 1.
}
if ( s1 == s3 ){
System.out.println("s1 and s3 are same" ); // 2.
}
if ( s1 == s4 ){
System.out.println("s1 and s4 are same" ); // 3.
}
if ( s1 == s5 ){
System.out.println("s1 and s5 are same" ); // 4.
}
will return:
s1 and s2 are same
s1 and s3 are same
s1 and s5 are same
In all the cases besides of s4 variable, a value for which was explicitly created using new operator and where intern method was not used on it's result, it is a single immutable instance that's being returned JVM's string constant pool.
Refer to JavaTechniques "String Equality and Interning" for more information.
On a recent project, some huge data structures were set up with data that was read in from a database (and hence not String constants/literals) but with a huge amount of duplication. It was a banking application, and things like the names of a modest set (maybe 100 or 200) corporations appeared all over the place. The data structures were already large, and if all those corp names had been unique objects they would have overflowed memory. Instead, all the data structures had references to the same 100 or 200 String objects, thus saving lots of space.
Another small advantage of interned Strings is that == can be used (successfully!) to compare Strings if all involved strings are guaranteed to be interned. Apart from the leaner syntax, this is also a performance enhancement. But as others have pointed out, doing this harbors a great risk of introducing programming errors, so this should be done only as a desparate measure of last resort.
The downside is that interning a String takes more time than simply throwing it on the heap, and that the space for interned Strings may be limited, depending on the Java implementation. It's best done when you're dealing with a known reasonable number of Strings with many duplications.
I want to add my 2 cents on using == with interned strings.
The first thing String.equals does is this==object.
So although there is some miniscule performance gain ( you are not calling a method), from the maintainer point of view using == is a nightmare, because some interned strings have a tendency to become non-interned.
So I suggest not to rely on special case of == for interned strings, but always use equals as Gosling intended.
EDIT: interned becoming non-interned:
V1.0
public class MyClass
{
private String reference_val;
...
private boolean hasReferenceVal ( final String[] strings )
{
for ( String s : strings )
{
if ( s == reference_val )
{
return true;
}
}
return false;
}
private void makeCall ( )
{
final String[] interned_strings = { ... init with interned values ... };
if ( hasReference( interned_strings ) )
{
...
}
}
}
In version 2.0 maintainer decided to make hasReferenceVal public, without going into much detail that it expects an array of interned strings.
V2.0
public class MyClass
{
private String reference_val;
...
public boolean hasReferenceVal ( final String[] strings )
{
for ( String s : strings )
{
if ( s == reference_val )
{
return true;
}
}
return false;
}
private void makeCall ( )
{
final String[] interned_strings = { ... init with interned values ... };
if ( hasReference( interned_strings ) )
{
...
}
}
}
Now you have a bug, that may be very hard to find, because in majority of cases array contains literal values, and sometimes a non-literal string is used. If equals were used instead of == then hasReferenceVal would have still continue to work. Once again, performance gain is miniscule, but maintenance cost is high.
Learn Java String Intern - once for all
Strings in java are immutable objects by design. Therefore, two string objects even with same value will be different objects by default. However, if we wish to save memory, we could indicate to use same memory by a concept called string intern. The below rules would help you understand the concept in clear terms:
String class maintains an intern-pool which is initially empty. This pool must guarantee to contain string objects with only unique values.
All string literals having same value must be considered same memory-location object because they have otherwise no notion of distinction. Therefore, all such literals with same value will make a single entry in the intern-pool and will refer to same memory location.
Concatenation of two or more literals is also a literal. (Therefore rule #2 will be applicable for them)
Each string created as object (i.e. by any other method except as literal) will have different memory locations and will not make any entry in the intern-pool
Concatenation of literals with non-literals will make a non-literal. Thus, the resultant object will have a new memory location and will NOT make an entry in the intern-pool.
Invoking intern method on a string object, either creates a new object that enters the intern-pool or return an existing object from the pool that has same value. The invocation on any object which is not in the intern-pool, does NOT move the object to the pool. It rather creates another object that enters the pool.
Example:
String s1=new String ("abc");
String s2=new String ("abc");
If (s1==s2) //would return false by rule #4
If ("abc" == "a"+"bc" ) //would return true by rules #2 and #3
If ("abc" == s1 ) //would return false by rules #1,2 and #4
If ("abc" == s1.intern() ) //would return true by rules #1,2,4 and #6
If ( s1 == s2.intern() ) //wound return false by rules #1,4, and #6
Note: The motivational cases for string intern are not discussed here. However, saving of memory will definitely be one of the primary objectives.
String literals and constants are interned by default.
That is, "foo" == "foo" (declared by the String literals), but new String("foo") != new String("foo").
you should make out two period time which are compile time and runtime time.for example:
//example 1
"test" == "test" // --> true
"test" == "te" + "st" // --> true
//example 2
"test" == "!test".substring(1) // --> false
"test" == "!test".substring(1).intern() // --> true
in the one hand,in the example 1,we find the results are all return true,because in the compile time,the jvm will put the "test" to the pool of literal strings,if the jvm find "test" exists,then it will use the exists one,in example 1,the "test" strings are all point to the same memory address,so the example 1 will return true.
in the other hand,in the example 2,the method of substring() execute in the runtime time,
in the case of "test" == "!test".substring(1),the pool will create two string object,"test"
and "!test",so they are different reference objects,so this case will return false,in the case of "test" == "!test".substring(1).intern(),the method of intern() will put the ""!test".substring(1)" to the pool of literal strings,so in this case,they are same reference objects,so will return true.
http://en.wikipedia.org/wiki/String_interning
string interning is a method of storing only one copy of each distinct string value, which must be immutable. Interning strings makes some string processing tasks more time- or space-efficient at the cost of requiring more time when the string is created or interned. The distinct values are stored in a string intern pool.
Interned Strings avoid duplicate Strings. Interning saves RAM at the expense of more CPU time to detect and replace duplicate Strings. There is only one copy of each String that has been interned, no matter how many references point to it. Since Strings are immutable, if two different methods incidentally use the same String, they can share a copy of the same String. The process of converting duplicated Strings to shared ones is called interning.String.intern() gives you the address of the canonical master String. You can compare interned Strings with simple == (which compares pointers) instead of equals which compares the characters of the String one by one. Because Strings are immutable, the intern process is free to further save space, for example, by not creating a separate String literal for "pot" when it exists as a substring of some other literal such as "hippopotamus".
To see more http://mindprod.com/jgloss/interned.html
String s1 = "Anish";
String s2 = "Anish";
String s3 = new String("Anish");
/*
* When the intern method is invoked, if the pool already contains a
* string equal to this String object as determined by the
* method, then the string from the pool is
* returned. Otherwise, this String object is added to the
* pool and a reference to this String object is returned.
*/
String s4 = new String("Anish").intern();
if (s1 == s2) {
System.out.println("s1 and s2 are same");
}
if (s1 == s3) {
System.out.println("s1 and s3 are same");
}
if (s1 == s4) {
System.out.println("s1 and s4 are same");
}
OUTPUT
s1 and s2 are same
s1 and s4 are same
String p1 = "example";
String p2 = "example";
String p3 = "example".intern();
String p4 = p2.intern();
String p5 = new String(p3);
String p6 = new String("example");
String p7 = p6.intern();
if (p1 == p2)
System.out.println("p1 and p2 are the same");
if (p1 == p3)
System.out.println("p1 and p3 are the same");
if (p1 == p4)
System.out.println("p1 and p4 are the same");
if (p1 == p5)
System.out.println("p1 and p5 are the same");
if (p1 == p6)
System.out.println("p1 and p6 are the same");
if (p1 == p6.intern())
System.out.println("p1 and p6 are the same when intern is used");
if (p1 == p7)
System.out.println("p1 and p7 are the same");
When two strings are created independently, intern() allows you to compare them and also it helps you in creating a reference in the string pool if the reference didn't exist before.
When you use String s = new String(hi), java creates a new instance of the string, but when you use String s = "hi", java checks if there is an instance of word "hi" in the code or not and if it exists, it just returns the reference.
Since comparing strings is based on reference, intern() helps in you creating a reference and allows you to compare the contents of the strings.
When you use intern() in the code, it clears of the space used by the string referring to the same object and just returns the reference of the already existing same object in memory.
But in case of p5 when you are using:
String p5 = new String(p3);
Only contents of p3 are copied and p5 is created newly. So it is not interned.
So the output will be:
p1 and p2 are the same
p1 and p3 are the same
p1 and p4 are the same
p1 and p6 are the same when intern is used
p1 and p7 are the same
public static void main(String[] args) {
// TODO Auto-generated method stub
String s1 = "test";
String s2 = new String("test");
System.out.println(s1==s2); //false
System.out.println(s1==s2.intern()); //true --> because this time compiler is checking from string constant pool.
}
string intern() method is used to create an exact copy of heap string object in string constant pool. The string objects in the string constant pool are automatically interned but string objects in heap are not. The main use of creating interns is to save the memory space and to perform faster comparison of string objects.
Source : What is string intern in java?
As you said, that string intern() method will first find from the String pool, if it finds, then it will return the object that points to that, or will add a new String into the pool.
String s1 = "Hello";
String s2 = "Hello";
String s3 = "Hello".intern();
String s4 = new String("Hello");
System.out.println(s1 == s2);//true
System.out.println(s1 == s3);//true
System.out.println(s1 == s4.intern());//true
The s1 and s2 are two objects pointing to the String pool "Hello", and using "Hello".intern() will find that s1 and s2. So "s1 == s3" returns true, as well as to the s3.intern().
By using heap object reference if we want to get corresponding string constant pool object reference, then we should go for intern()
String s1 = new String("Rakesh");
String s2 = s1.intern();
String s3 = "Rakesh";
System.out.println(s1 == s2); // false
System.out.println(s2 == s3); // true
Pictorial View
Step 1:
Object with data 'Rakesh' get created in heap and string constant pool. Also s1 is always pointing to heap object.
Step 2:
By using heap object reference s1, we are trying to get corresponding string constant pool object referenc s2, using intern()
Step 3:
Intentionally creating a object with data 'Rakesh' in string constant pool, referenced by name s3
As "==" operator meant for reference comparison.
Getting false for s1==s2
Getting true for s2==s3
Hope this help!!