if i create a string object as
String s=new String("Stackoverflow");
will String object created only in heap, or it also makes a copy in String constant pool.
Thanks in advance.
You only get a string into the constant pool if you call intern or use a string literal, as far as I'm aware.
Any time you call new String(...) you just get a regular new String object, regardless of which constructor overload you call.
In your case you're also ensuring that there is a string with contents "Stackoverflow" in the constant pool, by the fact that you're using the string literal at all - but that won't add another one if it's already there. So to split it up:
String x = "Stackoverflow"; // May or may not introduce a new string to the pool
String y = new String(x); // Just creates a regular object
Additionally, the result of a call to new String(...) will always be a different reference to all previous references - unlike the use of a string literal. For example:
String a = "Stackoverflow";
String b = "Stackoverflow";
String x = new String(a);
String y = new String(a);
System.out.println(a == b); // true due to constant pooling
System.out.println(x == y); // false; different objects
Finally, the exact timing of when a string is added to the constant pool has never been clear to me, nor has it mattered to me. I would guess it might be on class load (all the string constants used by that class, loaded immediately) but it could potentially be on a per-method basis. It's possible to find out for one particular implementation using intern(), but it's never been terribly important to me :)
In this case you are constructing an entirely new String object and that object won't be shared in the constant pool. Note though that in order to construct your new String() object you actually passed into it a String constant. That String constant is in the constants pool, however the string you created through new does not point to the same one, even though they have the same value.
If you did String s = "Stackoverflow" then s would contain a reference to the instance in the pool, also there are methods to let you add Strings to the pool after they have been created.
The new String is created in the heap, and NOT in the string pool.
If you want a newly create String to be in the string pool you need to intern() it; i.e.
String s = new String("Stackoverflow").intern();
... except of course that will return the string literal object that you started with!!
String s1 = "Stackoverflow";
String s2 = new String(s1);
String s3 = s2.intern();
System.out.println("s1 == s2 is " + (s1 == s2));
System.out.println("s2 == s3 is " + (s2 == s3));
System.out.println("s1 == s3 is " + (s1 == s3));
should print
s1 == s2 is false
s2 == s3 is false
s1 == s3 is true
And to be pedantic, the String in s is not the String that was created by the new String("StackOverflow") expression. What intern() does is to lookup the its target object in the string pool. If there is already a String in the pool that is equal(Object) to the object being looked up, that is what is returned as the result. In this case, we can guarantee that there will already be an object in the string pool; i.e. the String object that represents the value of the literal.
A regular java object will be created in the heap, and will have a reference s type of String. And, there will be String literal in String Constant Pool. Both are two different things.
My answer is YES!
Check the following code first:
String s0 = "Stackoverflow";
String s1 = new String("Stackoverflow");
String s2 = s1.intern();
System.out.println(s0 == s1);
System.out.println(s1 == s2 );
System.out.println(s0 == s2);
//OUTPUT:
false
false
true
s0 hold a reference in the string pool, while new String(String original) will always construct a new instance. intern() method of String will return a reference in the string pool with the same value.
Now go back to your question:
Will String object created only in heap, or it also makes a copy in String constant pool?
Since you already wrote a string constant "Stackoverflow" and pass it to your String constructor, so in my opinion, it has the same semantic as:
String s0 = "Stackoverflow";
String s1 = new String(s0);
which means there will be a copy in String constant pool when the code is evaluated.
But, if you construct the String object with following code:
String s = new String("StackoverflowSOMETHINGELSE".toCharArray(),0,13);
there won't be a copy of "Stackoverflow" in constant pool.
Related
This question already has answers here:
How do I compare strings in Java?
(23 answers)
Closed 6 years ago.
String s1 = "abc";
String s2 = "abc";
String s3 = new String("abc");
String s4 = new String("abc");
if (s1 == s2) is giving true
while (s3 == s4) is giving false.
Can somebody give a detailed explanation onto what is String pool, heap, how many objects are created in each line and how many objects created in total.
Why s3==s4 is giving false?
A detailed explanation will be much appreciated.
When you do new String(...), it is evaluated at runtime and hence creates two different instances and you end up getting s3 == s4 as false. Same thing happens when you use StringBuilder and do something like sb.toString() which is again evaluated at runtime.
For Example,
StringBuilder sb = new StringBuilder();
String foo = sb.append("abc").toString();
String bar = new String("abc");
String foobar = "abc";
Here foo, bar and foobar are all different objects and hence foo == bar or foo == foobar will evaluate to false.
On the other hand s1 == s2 returns true because abc one declared, already exists in the String Pool and in this case both the objects points to the same reference in the pool.
String Class in java is defined in java.lang package and it is exactly that, a class and not a primitive like int or boolean.
Strings are developed to offer operations with many characters ans are commmonly used in almost all the Java applications
Some interesting facts about Java and Strings:
String in immutable and final in Java and in this case JVM uses String Pool to store all the String objects.
What are different ways to create String Object?
We can create String object using new operator like any normal java class or we can use double quotes (literal assignment) to create a String object.
There are too several constructors available in String class to get String from char array, byte array, StringBuffer and StringBuilderetc etc.
To your Question:
When we create a String using double quotes, JVM looks in the String pool to find if any other String is stored with same value. If found, it just returns the reference to that String object else it creates a new String object with given value and stores it in the String pool.
When we use new operator, JVM creates the String object but don’t store it into the String Pool. We can use intern() method to store the String object into String pool or return the reference if there is already a String with equal value present in the pool.
So when you do
String s1 = "abc";
String s2 = "abc";
those are checked in the StringPool and since s1 already exist there, s2 will take the same reference, hence, s1 ==s2 is true.
but when you do:
String s3 = new String("abc");
String s4 = new String("abc");
you are using the new operator, therefore the JVM is not checking if there is an string already in the heap, it will just allocate a new space for s4, so is s3==s4 ??? of course no.
Please take a look at the image below for a more illustrative example.
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!!
How many distinct String object instances are created in the following code segment?
String s1 = new String("hello");
String s2 = "GoodBye";
String s3 = s1;
Not sure about all my reasoning here.
By using the keyword new that creates an instance from the String class, I am guessing that has to be an object. However, I am confused, is String after the new now considered a method because it has ( ) and then it is calling a String literal "hello" in it?
String s2 = "Goodbye";
I think this is a String literal, and since Strings are actually objects even the String literal is considered object. Not 100% sure if that is true.
String s3 = s1; Just refers back to s1. Therefore, it is not distinct.
So my answer is 2 distinct objects.
Please explain if I am a right or wrong.
The correct answer is 3.
String s1 = new String("hello");
String s2 = "GoodBye";
String s3 = s1;
The compiler will put both literals "hello" and "GoodBye" into a "constant pool" during compilation time, which then will be loaded by the classloader. So the JVM automatically interns all String literals used by this class, when it loads that class. More about this: When are Java Strings interned?. The String constant pool is then managed during runtime.
During runtime the JVM will create the third String object when it reaches the line String s1 = new String("hello").
So you would and up with three distinct String objects, where two of them contain the same word "hello". So s1.equals("hello") would be true, but s1 == "hello" would be false, because s1 references to a different String on the heap, than the literal "hello".
The line String s3 = s1 just creates a variable s3 with a copied reference to the String object of s1. It doesn't create a new object.
Also mind that you can "manually" add Strings into the String constant pool by using the method String#intern. So s1.intern() == "hello" is true, because the String reference returned from s1.intern() is the reference to the literal "hello" which was already in the constant pool.
If you like to get another and maybe more detailed explanation with some drawings about objects and their location, you can check this article on javaranch.
We use intern() method for the Strings which are created through new String() so that they will create an entry in String Pool and return the newly created String from the String Pool so that this created string is eligible for use with == operator (as per my understanding).
Then what is the use of creating a new String through constructor?
When should we use constructor for creating new String?
new String() can create at-most two Objects of String and at-least one. one in constant pool (if same is not present in constant pool) and another in heap memory.
constant pool entry is created by interning String object.
intern will create String Literal in Constant pool so that whenever you create String without new keyword it will not create string object.
suppose you have created as
String str= new String("abc");
two object will be created one in constant pool (if "abc" is not present in constant pool). jvm wil internally call intern for that object.
so next time if you are doing
String str1= "abc";
no entry will be added to constant pool because only entry can be possible in constant pool for same literal.
public native String intern(); java doc says,
A pool of strings, initially empty, is maintained privately by the class >{#code String}. When the intern method is invoked, if the pool already >contains a string equal to this {#code String} object as determined by >the {#link #equals(Object)} method, then the string from the pool is >returned. Otherwise, this {#code String} object is added to the pool and >a reference to this {#code String} object is returned. It follows that >for any two strings {#code s} and {#code t}, {#code s.intern() == >t.intern()} is {#code true} if and only if {#code s.equals(t)} is {#code >true}.
Let's consider an example:
String temp1 = new String("abcd"); this means, a new object "abcd" will be created in the heap memory and its reference will be linked to temp1.
String temp2 = new String("abcd").intern(); this means "abcd" literal will be checked in the String Pool. If already present, its reference will be linked to newly created temp2. Else a new entry will be created in String Pool and its reference will be linked to newly created temp2.
String temp3 = "abcd"; this means "abcd" literal will be checked in the String Pool. If already present, its reference will be linked to newly created temp2. Else a new entry will be created in String Pool and its reference will be linked to newly created temp3. This conclude that, Java automatically interns String literals.
Let's consider another example:
String s1 = "string-test";
String s2 = new String("string-test");
String s3 = new String("string-test").intern();
if ( s1 == s2 ){
System.out.println("s1 and s2 are same");
}
if ( s2 == s3 ){
System.out.println("s2 and s3 are same" );
}
if ( s1 == s3 ){
System.out.println("s1 and s3 are same" );
}
Output: s1 and s3 are same
Now when to create a String object using = " " or using new String(" ")?
One use case which I came across,
// imagine a multi-megabyte string here
String s = "0123456789012345678901234567890123456789";
String s2 = s.substring(0, 1);
s = null;
You'll now have a String s2 which, although it seems to be a one-character string, holds a reference to the gigantic char array created in the String s. This means the array won't be garbage collected, even though we've explicitly nulled out the String s!
The fix for this is to use String constructor like this:
String s2 = new String(s.substring(0, 1));
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!!