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Questions about Java's String pool [duplicate]
(7 answers)
Closed last month.
If I write something like this:
String s1 = new String("ABC");
String s2 = "ABC";
In which scenarios does the string pool get updated? The first, second or both?
In the above programming code ,when string pool get updated?
The first statement will update/add to the String pool with "ABC" String literal, but create another object in the heap and the variable s1 will refer the heap object.
The second statement will refer the already created Pool String object.
String s = new String("ABC");
this always create a new String object on the heap and add "ABC" to the pool(if not present).
String s= "ABC";
Whereas this line is called string literal. It checks whether the string pool already has the same string "ABC" or not. If present then s will refer to that object otherwise a new object will be created.
Conclusion: new String() will always create a new object and string literal checks the string pool before creating one.
String s1=new String("ABC");//creates two objects and one reference variable
In your case, JVM will create a new String object in normal(nonpool) Heap memory and the literal "ABC" will be placed in the string constant pool.The variable s1 will refer to the object in Heap(nonpool).
String Constant Pool
String objects created with the new operator do not refer to objects
in the string pool but can be made to using String’s intern() method.
The java.lang.String.intern() returns an interned String, that is, one
that has an entry in the global String literal pool. If the String is
not already in the global String literal pool, then it will be added.
Whenever you use new keyword to create String Object,it will create in Heap and then it will check the String Constant Pool for the same String Literal.If SCP doesn't contain that String literal then Only it will create String literal in the SCP.
Nope. Strings created with the constructor never enter the string pool.
In the example, two separate objects are created. Therefore:
s1 != s2
Related
Well, I've got such a code
String s = "hello";
String s2 = s + "world";
I know that variable s is stored in Java Heap in String Literal Pool. Also I know that variable s2 is stored in Java Heap (because '+' in this example creates a new object). But my question is: will "world" be put into Pool or nothing will happen with it?
Yes it will. Wherever / Whenever , you have a String literal declared, it goes to pool obviously, regardless the usage of it.
Since the String "world" does not exist in the pool, a new String object instantiates, then it is placed in the pool.
When we use double quotes to create a String, it first looks for String with same value in the String pool, if found it just returns the reference else it creates a new String in the pool and then returns the reference.
Basically, 's' and 's2' are just references while real objects are "Hello" and "world". This follows the concept specifying Strings are immutable while String references are not. So, Yes, as "world" is String object,apart from being a String literal, it will be stored in heap similar to hello. Also remember neither 's' nor 's2' will goto heap as being mutable String references.
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When should we use intern method of String on String literals
(14 answers)
Closed 7 years ago.
We know that All string literals in Java programs, such as "abc", are implemented as instances of String class and all this goes to String Constant Pool.
my question is : where the "Hello" literal will store in below case?
String str=new String("Hello");
if this literal will goes to "String Constant Pool" then what intern method will do?
String literals in your code go directly to the Sring pool. Calling intern() on them will do nothing. intern() will only make sense in cases where you are dinamically constructing Strings (in runtime, not compile-time). In those cases calling intern() will instruct the JVM to move this variable into the String pool for future usage.
It will store into Heap. If you will use intern method then it will be pushed to String pool.
public static void main(String[] args) {
String strLiteral="Hello";
String str=new String("Hello");
System.out.println(str==strLiteral);
String str2=new String("Hello").intern();
System.out.println(str2==strLiteral);
}
Output:-
false
true
1) Look at the code above, I have created a String Literal Hello, which is stored in String Pool.
2) Now I have created a String object str, if I check the references str and strLiteral they are not equal i.e. they are pointing to two different objects one in heap and other one in String pool.
3) Now I have created a String object str2 and called an intern method on it, JVM now will search the for "Hello" in String pool and return the reference to it. So now if I am checking strLiteral and str2 are equal i.e. pointing to the same object in String pool.
String s="abc" in java.what happened in the memory any object is created or not and what is "s" here variable or object,and same question with String s=new String("abc");.
There's a thing called String Memory Pool in java, when you declare:
String str1="abc";
It goes to that memory pool and not on the heap. But when you write:
String str2=new String("abc");
It creates a full fledged object on the heap, If you again write:
String str3 = "abc";
It won't create any more object on the pool, it will check the pool if this literal already exists it will assign that to it. But writing:
String str4 = new String("abc");
will again create a new object on the heap
Key point is that:
A new object will always be created on the heap as many times as you keep writing:
new String("abc");
But if you keep assigning the Strings directly without using the keyword new, it will just get referenced from the pool (if it exists in the pool)
The String class represents character strings. All string literals in Java programs, such as "abc", are implemented as instances of this class.
String Java Doc
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Strings are objects in Java, so why don't we use 'new' to create them?
(15 answers)
Closed 9 years ago.
I was reading some advices about best practices in java and I got the following idea which made me curious
Also whenever you want to instantiate a String object, never use its constructor but always instantiate it directly.
For example:
//slow instantiation
String slow = new String("Yet another string object");
//fast instantiation
String fast = "Yet another string object";
Why is this? doesn't the 'fast' call the default string constructor?
When you use new you get a new string object, but if you use string literal then see here:
In computer science, 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. The single copy of each string is called its 'intern' and is
typically looked up by a method of the string class, for example
String.intern() in Java. All compile-time constant strings in Java are
automatically interned using this method.
If you do:
String a = "foo";
String b = "foo";
Then a==b is true !
A String will be created only if it hasn't been interned.
An object will be created in the first time, and it'll be stored in a place called the String constant pool.
But using the new which will create a different object for each string, will output false.
String a = new String("foo");
String b = new String("foo");
Now a==b is false.
So when using literal it is easier to read, plus easier for the compiler to make optimizations. So.. use it when you can.
The JVM maintains a pool of String literals for optimizations. When you create a String using the constructor,
String s1 = new String("foo");
A new String object is created, and the literal "foo" is added to the pool. After this, anytime you use "foo" in your code, the "foo" refers to the item in the pool and a new object is not created. Since String is immutable, this does not create any problems.
So when you create a String using the "shortcut":
String s2 = "foo"
the JVM looks into the pool, if "foo" already exists there, it will make s2 refer to the item in the pool.
This is a major difference with possible performance impacts: The constructor always creates an object, and adds the literal to the pool if it is not already present there. The shortcut refers to the item in the pool and creates a new object only if the liuteral is not in the pool.
This question already has answers here:
Questions about Java's String pool [duplicate]
(7 answers)
Closed 5 years ago.
When we use new operator to create a String object, I read that two Objects are created one Object is string constant pool and second one is in heap memory.
My question here is We are using new operator hence only one object should be created in Heap. Why then one more object has to be created in String Constant pool. I know Java stores String object whenever we not use new operator to create a String. For eg:
String s = "abc" .
In this case only it will create in String constant pool.
String s2 = new String("abc")
only one object hast to be created in heap and not in Constant pool.
Please explain why I am wrong here.
We are using new operator hence only one object should be created in Heap.
Sure - the new operation only creates one object. But its parameter is a String literal, which already represents an object. Any time you use a String literal, an object was created for that during class loading (unless the same literal was already used elsewhere). This isn't skipped just because you then use the object as a parameter for a new String() operation.
And because of that, the new String() operation is unnecessary most of the time and rarely used.
See http://docs.oracle.com/javase/specs/jls/se7/html/jls-15.html#jls-15.28
Compile-time constant expressions of type String are always "interned" so as to share unique instances, using the method String.intern
Thus when you write
String s2 = new String("abc")
The Compile-time constant expression "abc" will be interend.
It will be created two object, one object into pool area and other one in non pool area because you are using new and as well as string literal as a parameter.
String s = new String("abs");
String s2 = new String("abc")
It will create only one object in heap memory. We need to use intern() method of java.lang.String explicitly to make the entry in String pool.
String s = "def".
Two Objects will be created here. When you create using String literal notation of Java, it automatically call intern() method to put that object into String pool, provided it was not present in the pool already.
String in java is Immutable, once created its cannot be changed. also, any string literal will be stored in the string pool for later use whenever the same literal is used again, for example:
String name = "my name"; // one intern object is created in pool
String otherName = "my name"; // the old intern object is reused
System.out.println( name == otherName); // true , the same reference refer to same object
the two reference refer to the same location in the pool.
String name = new String("my name"); // one object is created, no string pool checking
String otherName = new String("my name"); // other object is created, no string pool checking
System.out.println( name == otherName); // false, we have 2 different object
here we have two different String object in memory each of them have its own value.
see this article for more:
http://www.javaranch.com/journal/200409/Journal200409.jsp#a1
form this below line of code will create 3 object in the JVM
String s2 = new String("abc")
for "abc" in String pool memory.
for new operator one object in heap.
one more for s2.
String s2 = new String("abc");
here 1 literal and 1 object will be created.
With new operator 1 String object will be created in heap as "abc" and s2 will refer it, moreover "abc" is string literal also that is passed in String Constructor so it will go in String Constant pool.
literal is consider as object so 2 object will be created here.