Consider statement:
String s=new String("abc");
Will this statement creates two String objects namely "abc" and the one represented by 's'?
and if it creates two objects then will "abc" get stored in String pool or just discarded?
EDIT:
i am asking this question in reference to Difference between string object and string literal, where in the last two answers , creation of two objects is denied.
Avoid such kind of behavior , because "abc" is already a String and by making a new String, you are creating an unnecessary Object.
Instead go for String s = "abc";
This way, the String gets interned by the JVM and is added to a pool.
To answer your question, you are just creating an Object s that is referring to "abc".
So when you do say String t = new String("abc"); and then do s==t, will yield in false. Because they have their separate instances to abc.
String s = "HELLO";
Here "s" is a object reference variable of type String, which refers to the String literal object "Hello" which is added to the String Literal Pool.
String t = new String("Hello");
Here t is a object reference variable of type String, which refers to the String object "Hello" which is added to the String Pool.
Difference Between String Literal and String :
Assume
String s = "Hello";
String t = new String("Hello");
Now if following changes are done:
s = null;
t = null;
Hello String object associated with t will be a candidate for Garbage Collector, But Hello String Literal associated with s will NOT BE A CANDIDATE for Garbage Collector, as there will ALWAYS BE A REFERENCE FROM STRING LITERAL POOL to it.
Related
As we know that when we create new String object by new keyword like this:
String str = new String("New String Will Have Two Objects");
It will create two objects of, one on java heap memory and other on String pool.
So when we call access "str" which string object is accessed(heap object or string pool object)?
According to my understanding the string pool object is get accessed, if yes then what happens to heap object?
If you are creating the String object with new
String str = new String("New String Will Have Two Objects");
In such case, JVM will create a new string object in normal(non pool) heap memory and the literal "New String Will Have Two Objects" will be placed in the string constant pool. The variable str will refer to the object in heap(non pool).
Method ‘intern()’ usage
This is best described by java docs
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.
String str = new String("New String Will Have Two Objects");
str.intern();
new String creates a new string on heap. This string's reference gets assigned to str.
String literals always have a reference to them from the String Literal Pool. That means that they always have a reference to them and are, therefore, not eligible for garbage collection http://www.javaranch.com/journal/200409/ScjpTipLine-StringsLiterally.html
Let's say that I need to iteratively retrieve a value of the same key from a Java hashmap.
for(int i=0; i<INTEGER.MAX; i++)
map.get("KEY");
In this case, is the "KEY" string created every time I call map.get("KEY")? I was wondering if it's always better to have a String constant, or it doesn't matter.
No. String constants are interned automatically, so any identical string literals all reference the same object in memory.
Some more information on this: http://www.xyzws.com/Javafaq/what-is-string-literal-pool/3
An example of this:
String s1 = "Test";
String s2 = "Test";
String s3 = new String("Test");
s1 == s2;//Evaluates to true, because they are the same object (both created with string literals)
s1 == s3;//Evaluates to false, because they are different objects containing identical data
Yes/No Answer depends on how you create String Objects. Below are the four scenarios I can think of as of now.
Yes Cases
new String() always creates new Object. It is not internedn(Doesn't go to String pool) so you
can not take it back from memory.
Concatenation ( "a" + "b" ) always creates new String Object and it is not interned (Doesn't go to String pool).
No Cases
String a ="aa"; if already available it retrieves from the pool, when not available it creates a new object which is interned also (Goes to String pool as well)
new String().intern() or "aa".intern(); if already available it retrieves from pool , when not available it creates new object which
is interned also (Goes to String pool as well).
is the "KEY" string created every time I call map.get("KEY")?
No.
Java Strings are immutable, which allows the Java compiler to use a single instance for all string literals.
That is: all identical string literals in your program will reference a single string object.
In the rare cases you need identical strings to be wrapped in two separate objects, you must explicitly
instantiate a String object:
String s1 = "bla";
String s2 = "bla";
// s1 == s2
String s3 = new String ("bla");
// s1 != s3
I recently learned that in Java: == compares the object references, not the content, which is why:
String str1 = "hell";
String str2 = "o";
String str3 = str1 + str2;
String str4 = "hello";
str3 == str4; // False
So far so good. However when I do the following:
String str5 = "hello";
str5 == str4; // True
Does this mean that str5 and str4 reference the same memory object? How does this work?
The String str5 = "hello"; creates a pooled String value hello, which is why str5 == str4 returns true.
On the other hand, str1 + str2 works like this:
An instance of the StringBuilder class is created (behind the scenes)
The + operator actually invokes the StringBuilder#append(String s) method
When the appending is done, a StringBuilder.toString() method is invoked, which returns a brand new String object. This is why str3 == str4 is actually false.
More info:
How do I compare Strings in Java?
How Java do the string concatenation using “+”?
Yes. str5 and str4 refer the same memory object. As Strings are immutable when you change the value of some string its produce an different object. If two String objects have the same value then second one is not created, JVM just give the reference of the first object.
When the value of String changed different object created for some security and other usefull purpose read these link:
Immutability of Strings
wiki Immutable object
When some string are created like
String str1="hello";
JVM creates an immutable object when again you try to create some string with same value
String str2="hello"
JVM use the same procedure to create an object as see's that this object is already created then its return the object of the str1 to reduce duplicate object creation.
This will be useful string pool in the jvm
Yes, when you create and assign a String value eg String s1="hello"; , it gets added in the String pool. Now if you assign the same String value to another reference like this:-
String s2="hello";
The variable s2 will point to the same String object hello , present in String pool.
However you can force and create a new String object for the same values like this:-
String s3= new String("hello");
This will add and create new object for hello even though it is already present in the String pool.
Hence it can be summarised as:-
s1==s2; //return true
s1==s3; //return false
s2==s3; //returns false
Strings are pooled by the JVM once they are created, that's why those variables refer to the same instance in the String pool.
And the idea behind having a String pool is to avoid unnecessary object creation.
Strings are immutable objects. This is why str1 and str2 combined are not equal to str3.
Ya,its gives true for str5 ==str4 becoz it uses"string pool area" to store."this time also it
compares object references" but these strings has same object reference as string pool area has one object id.
In first case strings are not created in string pool area thats why it gives false.
A private pool of string literals is maintained by String class. All literal strings and string-valued constant expressions are added to this pool during program start up. So all string literals with same value will point to same object.
When a new string object is created(e.g. by concatenating two string objects) it does not belong to the String pool, so comparing it with another string literal will return false. Any string object can be added to String pool by invoking intern() method on that string object, now the object will point to a string literal from the pool which have same value as this object. Now any string comparison of this object with same string literal will yield true.
I know there are two ways of creating String in Java:
String a = "aaa";
String b = new String("bbb");
With the first way Java will definitely create a String object in the string pool and make a refer to it. (Assume "aaa" wan't in the pool before.)
With the second method, an object will be created in the heap, but will jvm also create an object in the string pool?
In this post Questions about Java's String pool, #Jesper said:
If you do this:
String s = new String("abc");
then there will be one String object in the pool, the one that represents the literal "abc", > and there will be a separate String object, not in the pool, that contains a copy of the > content of the pooled object.
If that's true, then every time with the new String("bbb");, a object "bbb" is created in the pool, which means by either way above, java will always create a string object in the pool. Then what is intern() used for ? In the docs http://docs.oracle.com/javase/7/docs/api/java/lang/String.html#intern(), it says:
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.
That means there are cases that a string is not in the pool, is that possible ? Which one is true ?
As you know that String is an immutable object in Java programming language, which means once constructed can not be altered. Due to this, JVM has the ability to maintain a literal pool which is helpful to reduce the memory usage and to increase the performance. Each time when a String literal is used JVM checks the literal pool. If the literal is already available, the same reference would be returned. If the literal is not available, a new String object will be created and added in the literal pool.
This theory is applied when you try to create a String like a primitive or a literal/constant.
String str = "bbb";
But when you create a new String object
String str = new String("bbb");
the above mentioned rules are overridden and a new instance is created always.
But the intern API in the String class can be used to pick the String reference from the literal pool even though you create a String using new operator. Please check the below given example. Although the str3 is created using new operator since we used the intern method JVM picked up the reference from the literal pool.
public class StringInternExample {
public static void main(final String args[]) {
final String str = "bbb";
final String str1 = "bbb";
final String str2 = new String("bbb");
final String str3 = new String("bbb").intern();
System.out.println("str == str1 : "+(str == str1));
System.out.println("str == str2 : "+(str == str2));
System.out.println("str == str3 : "+(str == str3));
}
}
Output of above code:
str == str1 : true
str == str2 : false
str == str3 : true
You can have a look: Confusion on string immutability
Source of answer: http://ourownjava.com/java/java-string-immutability-and-intern-method/
Shishir
There are essentially two ways that our String objects can enter in to the pool:
Using a literal in source code like "bbb".
Using intern.
intern is for when you have a String that's not otherwise from the pool. For example:
String bb = "bbb".substring(1); // substring creates a new object
System.out.println(bb == "bb"); // false
System.out.println(bb.intern() == "bb"); // true
Or slightly different:
System.out.println(new String("bbb").intern() == "bbb"); // true
new String("bbb") does create two objects...
String fromLiteral = "bbb"; // in pool
String fromNewString = new String(fromLiteral); // not in pool
...but it's more like a special case. It creates two objects because "bbb" refers to an object:
A string literal is a reference to an instance of class String [...].
Moreover, a string literal always refers to the same instance of class String.
And new String(...) creates a copy of it.
However, there are many ways String objects are created without using a literal, such as:
All the String methods that perform some kind of mutation. (substring, split, replace, etc.)
Reading a String from some kind of input such as a Scanner or Reader.
Concatenation when at least one operand is not a compile-time constant.
intern lets you add them to the pool or retrieve an existing object if there was one. Under most circumstances interning Strings is unnecessary but it can be used as an optimization because:
It lets you compare with ==.
It can save memory because duplicates can be garbage collected.
Yes, new String("abc") will create a new object in memory, and thus it is advised to avoid it. Please have a look at item 5 of Josh Bloch's Effective Java, "Avoid creating unnecessary objects" where it is better explained:
As an extreme example of what not to do, consider this statement:
String s = new String("stringette"); // DON'T DO THIS!
The statement
creates a new String instance each time it is executed, and none of
those object creations is necessary. The argument to the String
constructor ("stringette") is itself a String instance, functionally
identical to all of the objects created by the constructor. If this
usage occurs in a loop or in a frequently invoked method, millions of
String instances can be created needlessly. The improved version is
simply the following:
String s = "stringette";
This version uses a
single String instance, rather than creating a new one each time it is
executed. Furthermore, it is guaranteed that the object will be reused
by any other code running in the same virtual machine that happens to
contain the same string literal [JLS, 3.10.5].
http://uet.vnu.edu.vn/~chauttm/e-books/java/Effective.Java.2nd.Edition.May.2008.3000th.Release.pdf
With the second method, an object will be created in the heap, but will jvm also create an object in the string pool?
Yes, but it is the string literal "bbb" which ensures the interned string1. The string constructor creates a new string object which is a copy with the same length and content - the newly created string is not automatically interned.
If that's true, then every time with the new String("bbb");, a object "bbb" is created in the pool, which means by either way above, java will always create a string object in the pool. Then what is intern() used for ?
Only string literals are automatically interned. Other string objects must be manually interned, if such is the desired behavior.
That means there are cases that a string is not in the pool, is that possible ?
With the exception of manual calls to String.intern, only string literals result in interned strings.
While I would recommend using a specialized collection for such cases, interning may be useful where it can be used to avoid creating extra duplicate objects. Some use-cases where interning can be beneficial - as in, the same string value can appear many times - is in JSON keys and XML element/attribute names.
1 This is trivial to reason, consider:
String _b = "bbb"; // string from string literal (this is interned)
String b = new String(_b); // create a NEW string via "copy constructor"
b == _b // -> false (new did NOT return an interned string)
b.equals(_b) // -> true (but it did return an equivalent string)
b.intern() == _b // -> true (which interns to .. the same string object)
I read in Kathy Sierra book that when we create String using new operator like String s = new String("abc") In this case, because we used the new keyword, Java will create a new String object in normal (nonpool) memory, and s will refer to it. In addition, literal "abc" will be placed in the pool.
intern() says that if String pool already contains a string then the string from the pool is returned Otherwise, the String object is added to the pool and a reference to this String object is returned.
If string "abc" when created using new also placed the string in the pool, then wht does intern() says that string from the pool is returned if String pool contains the string otherwise the string object is added to the pool.
Also I want to know if we create a String using new then actually how many objects get created?
TL;DR: If you ever really need to do new String("abc"), you'll know you need to and you'll know why. It's so rare that it's almost valid to say you never need to. Just use "abc".
The long version:
When you have the code new String("abc") the following things occur at various times:
When the class containing that code is loaded, if a string with the characters "abc" is not already in the intern pool, it's created and put there.
When the new String("abc") code is run:
A reference to the "abc" string from the intern pool is passed into the String constructor.
A new String object is created and initialized by copying the characters from the String passed into the constructor.
The new String object is returned to you.
If string "abc" when created using new also placed the string in the pool, then why does intern() says that string from the pool is returned if String pool contains the string otherwise the string object is added to the pool.
Because that's what intern does. Note that calling intern on a string literal is a no-op; string literals are all interned automatically. E.g.:
String s1 = "abc"; // Get a reference to the string defined by the literal
String s2 = s1.intern(); // No-op
System.out.println(s1 == s2); // "true"
System.out.println(s1 == "abc"); // "true", all literals are interned automatically
Also I want to know if we create a String using new then actually how many objects get created?
You create at least one String object (the new, non-interned one), and possibly two (if the literal wasn't already in the pool; but again, that bit happens earlier, when the class file's literals are loaded):
String s1 = "abc"; // Get a reference to the string defined by the literal
String s2 = new String(s1); // Create a new `String` object (guaranteed)
System.out.println(s1 == s2); // "false"
String s3 = s2.intern(); // Get the interned version of the string with these characters
System.out.println(s1 == s3); // "true"
String Pool is a pool of string references. Objects are created in Heap only.
When using new String("abc").intern() or using method like String s = "abc"; String pool is checked if there is an reference existing which refers to "abc".
In case reference for "abc" already exists in pool and .intern() is called on the reference referencing to an String object created using new String("abc"), then object created by new String("abc") is eligible for garbage collection. See below code for more clarity.
public static void main(String[] args) {
String s = new String("abc");
String a = s;
System.out.println(s==a);// true
String b = "abc";
s = s.intern();
System.out.println(s==a);// false
}