What exactly does immutable mean - that is, what are the consequences of an object being mutable or immutable? In particular, why are Java's Strings immutable?
My understanding is that the StringBuilder type is something like a mutable equivalent to String. When would I use StringBuilder rather than String, and vice-versa?
Immutable means that once the constructor for an object has completed execution that instance can't be altered.
This is useful as it means you can pass references to the object around, without worrying that someone else is going to change its contents. Especially when dealing with concurrency, there are no locking issues with objects that never change
e.g.
class Foo
{
private final String myvar;
public Foo(final String initialValue)
{
this.myvar = initialValue;
}
public String getValue()
{
return this.myvar;
}
}
Foo doesn't have to worry that the caller to getValue() might change the text in the string.
If you imagine a similar class to Foo, but with a StringBuilder rather than a String as a member, you can see that a caller to getValue() would be able to alter the StringBuilder attribute of a Foo instance.
Also beware of the different kinds of immutability you might find: Eric Lippert wrote a blog article about this. Basically you can have objects whose interface is immutable but behind the scenes actual mutables private state (and therefore can't be shared safely between threads).
An immutable object is an object where the internal fields (or at least, all the internal fields that affect its external behavior) cannot be changed.
There are a lot of advantages to immutable strings:
Performance: Take the following operation:
String substring = fullstring.substring(x,y);
The underlying C for the substring() method is probably something like this:
// Assume string is stored like this:
struct String { char* characters; unsigned int length; };
// Passing pointers because Java is pass-by-reference
struct String* substring(struct String* in, unsigned int begin, unsigned int end)
{
struct String* out = malloc(sizeof(struct String));
out->characters = in->characters + begin;
out->length = end - begin;
return out;
}
Note that none of the characters have to be copied! If the String object were mutable (the characters could change later) then you would have to copy all the characters, otherwise changes to characters in the substring would be reflected in the other string later.
Concurrency: If the internal structure of an immutable object is valid, it will always be valid. There's no chance that different threads can create an invalid state within that object. Hence, immutable objects are Thread Safe.
Garbage collection: It's much easier for the garbage collector to make logical decisions about immutable objects.
However, there are also downsides to immutability:
Performance: Wait, I thought you said performance was an upside of immutability! Well, it is sometimes, but not always. Take the following code:
foo = foo.substring(0,4) + "a" + foo.substring(5); // foo is a String
bar.replace(4,5,"a"); // bar is a StringBuilder
The two lines both replace the fourth character with the letter "a". Not only is the second piece of code more readable, it's faster. Look at how you would have to do the underlying code for foo. The substrings are easy, but now because there's already a character at space five and something else might be referencing foo, you can't just change it; you have to copy the whole string (of course some of this functionality is abstracted into functions in the real underlying C, but the point here is to show the code that gets executed all in one place).
struct String* concatenate(struct String* first, struct String* second)
{
struct String* new = malloc(sizeof(struct String));
new->length = first->length + second->length;
new->characters = malloc(new->length);
int i;
for(i = 0; i < first->length; i++)
new->characters[i] = first->characters[i];
for(; i - first->length < second->length; i++)
new->characters[i] = second->characters[i - first->length];
return new;
}
// The code that executes
struct String* astring;
char a = 'a';
astring->characters = &a;
astring->length = 1;
foo = concatenate(concatenate(slice(foo,0,4),astring),slice(foo,5,foo->length));
Note that concatenate gets called twice meaning that the entire string has to be looped through! Compare this to the C code for the bar operation:
bar->characters[4] = 'a';
The mutable string operation is obviously much faster.
In Conclusion: In most cases, you want an immutable string. But if you need to do a lot of appending and inserting into a string, you need the mutability for speed. If you want the concurrency safety and garbage collection benefits with it the key is to keep your mutable objects local to a method:
// This will have awful performance if you don't use mutable strings
String join(String[] strings, String separator)
{
StringBuilder mutable;
boolean first = true;
for(int i = 0; i < strings.length; i++)
{
if(first) first = false;
else mutable.append(separator);
mutable.append(strings[i]);
}
return mutable.toString();
}
Since the mutable object is a local reference, you don't have to worry about concurrency safety (only one thread ever touches it). And since it isn't referenced anywhere else, it is only allocated on the stack, so it is deallocated as soon as the function call is finished (you don't have to worry about garbage collection). And you get all the performance benefits of both mutability and immutability.
Actually String is not immutable if you use the wikipedia definition suggested above.
String's state does change post construction. Take a look at the hashcode() method. String caches the hashcode value in a local field but does not calculate it until the first call of hashcode(). This lazy evaluation of hashcode places String in an interesting position as an immutable object whose state changes, but it cannot be observed to have changed without using reflection.
So maybe the definition of immutable should be an object that cannot be observed to have changed.
If the state changes in an immutable object after it has been created but no-one can see it (without reflection) is the object still immutable?
Immutable objects are objects that can't be changed programmatically. They're especially good for multi-threaded environments or other environments where more than one process is able to alter (mutate) the values in an object.
Just to clarify, however, StringBuilder is actually a mutable object, not an immutable one. A regular java String is immutable (meaning that once it's been created you cannot change the underlying string without changing the object).
For example, let's say that I have a class called ColoredString that has a String value and a String color:
public class ColoredString {
private String color;
private String string;
public ColoredString(String color, String string) {
this.color = color;
this.string = string;
}
public String getColor() { return this.color; }
public String getString() { return this.string; }
public void setColor(String newColor) {
this.color = newColor;
}
}
In this example, the ColoredString is said to be mutable because you can change (mutate) one of its key properties without creating a new ColoredString class. The reason why this may be bad is, for example, let's say you have a GUI application which has multiple threads and you are using ColoredStrings to print data to the window. If you have an instance of ColoredString which was created as
new ColoredString("Blue", "This is a blue string!");
Then you would expect the string to always be "Blue". If another thread, however, got ahold of this instance and called
blueString.setColor("Red");
You would suddenly, and probably unexpectedly, now have a "Red" string when you wanted a "Blue" one. Because of this, immutable objects are almost always preferred when passing instances of objects around. When you have a case where mutable objects are really necessary, then you would typically guard the objet by only passing copies out from your specific field of control.
To recap, in Java, java.lang.String is an immutable object (it cannot be changed once it's created) and java.lang.StringBuilder is a mutable object because it can be changed without creating a new instance.
In large applications its common for string literals to occupy large bits of memory. So to efficiently handle the memory, the JVM allocates an area called "String constant pool".(Note that in memory even an unreferenced String carries around a char[], an int for its length, and another for its hashCode. For a number, by contrast, a maximum of eight immediate bytes is required)
When complier comes across a String literal it checks the pool to see if there is an identical literal already present. And if one is found, the reference to the new literal is directed to the existing String, and no new 'String literal object' is created(the existing String simply gets an additional reference).
Hence : String mutability saves memory...
But when any of the variables change value, Actually - it's only their reference that's changed, not the value in memory(hence it will not affect the other variables referencing it) as seen below....
String s1 = "Old string";
//s1 variable, refers to string in memory
reference | MEMORY |
variables | |
[s1] --------------->| "Old String" |
String s2 = s1;
//s2 refers to same string as s1
| |
[s1] --------------->| "Old String" |
[s2] ------------------------^
s1 = "New String";
//s1 deletes reference to old string and points to the newly created one
[s1] -----|--------->| "New String" |
| | |
|~~~~~~~~~X| "Old String" |
[s2] ------------------------^
The original string 'in memory' didn't change, but the
reference variable was changed so that it refers to the new string.
And if we didn't have s2, "Old String" would still be in the memory but
we'll not be able to access it...
"immutable" means you cannot change value. If you have an instance of String class, any method you call which seems to modify the value, will actually create another String.
String foo = "Hello";
foo.substring(3);
<-- foo here still has the same value "Hello"
To preserve changes you should do something like this
foo = foo.sustring(3);
Immutable vs mutable can be funny when you work with collections. Think about what will happen if you use mutable object as a key for map and then change the value (tip: think about equals and hashCode).
java.time
It might be a bit late but in order to understand what an immutable object is, consider the following example from the new Java 8 Date and Time API (java.time). As you probably know all date objects from Java 8 are immutable so in the following example
LocalDate date = LocalDate.of(2014, 3, 18);
date.plusYears(2);
System.out.println(date);
Output:
2014-03-18
This prints the same year as the initial date because the plusYears(2) returns a new object so the old date is still unchanged because it's an immutable object. Once created you cannot further modify it and the date variable still points to it.
So, that code example should capture and use the new object instantiated and returned by that call to plusYears.
LocalDate date = LocalDate.of(2014, 3, 18);
LocalDate dateAfterTwoYears = date.plusYears(2);
date.toString()… 2014-03-18
dateAfterTwoYears.toString()… 2016-03-18
I really like the explaination from SCJP Sun Certified Programmer for Java 5 Study Guide.
To make Java more memory efficient, the JVM sets aside a special area of memory called the "String constant pool." When the compiler encounters a String literal, it checks the pool to see if an identical String already exists. If a match is found, the reference to the new literal is directed to the existing String, and no new String literal object is created.
Objects which are immutable can not have their state changed after they have been created.
There are three main reasons to use immutable objects whenever you can, all of which will help to reduce the number of bugs you introduce in your code:
It is much easier to reason about how your program works when you know that an object's state cannot be changed by another method
Immutable objects are automatically thread safe (assuming they are published safely) so will never be the cause of those hard-to-pin-down multithreading bugs
Immutable objects will always have the same Hash code, so they can be used as the keys in a HashMap (or similar). If the hash code of an element in a hash table was to change, the table entry would then effectively be lost, since attempts to find it in the table would end up looking in the wrong place. This is the main reason that String objects are immutable - they are frequently used as HashMap keys.
There are also some other optimisations you might be able to make in code when you know that the state of an object is immutable - caching the calculated hash, for example - but these are optimisations and therefore not nearly so interesting.
One meaning has to do with how the value is stored in the computer, For a .Net string for example, it means that the string in memory cannot be changed, When you think you're changing it, you are in fact creating a new string in memory and pointing the existing variable (which is just a pointer to the actual collection of characters somewhere else) to the new string.
String s1="Hi";
String s2=s1;
s1="Bye";
System.out.println(s2); //Hi (if String was mutable output would be: Bye)
System.out.println(s1); //Bye
s1="Hi" : an object s1 was created with "Hi" value in it.
s2=s1 : an object s2 is created with reference to s1 object.
s1="Bye" : the previous s1 object's value doesn't change because s1 has String type and String type is an immutable type, instead compiler create a new String object with "Bye" value and s1 referenced to it. here when we print s2 value, the result will be "Hi" not "Bye" because s2 referenced to previous s1 object which had "Hi" value.
Immutable means that once the object is created, non of its members will change. String is immutable since you can not change its content.
For example:
String s1 = " abc ";
String s2 = s1.trim();
In the code above, the string s1 did not change, another object (s2) was created using s1.
Immutable simply mean unchangeable or unmodifiable. Once string object is created its data or state can't be changed
Consider bellow example,
class Testimmutablestring{
public static void main(String args[]){
String s="Future";
s.concat(" World");//concat() method appends the string at the end
System.out.println(s);//will print Future because strings are immutable objects
}
}
Let's get idea considering bellow diagram,
In this diagram, you can see new object created as "Future World". But not change "Future".Because String is immutable. s, still refer to "Future". If you need to call "Future World",
String s="Future";
s=s.concat(" World");
System.out.println(s);//print Future World
Why are string objects immutable in java?
Because Java uses the concept of string literal. Suppose there are 5 reference variables, all refers to one object "Future".If one reference variable changes the value of the object, it will be affected to all the reference variables. That is why string objects are immutable in java.
Once instanciated, cannot be altered. Consider a class that an instance of might be used as the key for a hashtable or similar. Check out Java best practices.
As the accepted answer doesn't answer all the questions. I'm forced to give an answer after 11 years and 6 months.
Can somebody clarify what is meant by immutable?
Hope you meant immutable object (because we could think about immutable reference).
An object is immutable: iff once created, they always represent the same value (doesn't have any method that change the value).
Why is a String immutable?
Respect the above definition which could be checked by looking into the Sting.java source code.
What are the advantages/disadvantages of the immutable objects?
immutable types are :
safer from bugs.
easier to understand.
and more ready for change.
Why should a mutable object such as StringBuilder be preferred over String and vice-verse?
Narrowing the question Why do we need the mutable StringBuilder in programming?
A common use for it is to concatenate a large number of strings together, like this:
String s = "";
for (int i = 0; i < n; ++i) {
s = s + n;
}
Using immutable strings, this makes a lot of temporary copies — the first number of the string ("0") is actually copied n times in the course of building up the final string, the second number is copied n-1 times, and so on. It actually costs O(n2) time just to do all that copying, even though we only concatenated n elements.
StringBuilder is designed to minimize this copying. It uses a simple but clever internal data structure to avoid doing any copying at all until the very end, when you ask for the final String with a toString() call:
StringBuilder sb = new StringBuilder();
for (int i = 0; i < n; ++i) {
sb.append(String.valueOf(n));
}
String s = sb.toString();
Getting good performance is one reason why we use mutable objects. Another is convenient sharing: two parts of your program can communicate more conveniently by sharing a common mutable data structure.
More could be found here : https://web.mit.edu/6.005/www/fa15/classes/09-immutability/#useful_immutable_types
Immutable Objects
An object is considered immutable if its state cannot change after it is constructed. Maximum reliance on immutable objects is widely accepted as a sound strategy for creating simple, reliable code.
Immutable objects are particularly useful in concurrent applications. Since they cannot change state, they cannot be corrupted by thread interference or observed in an inconsistent state.
Programmers are often reluctant to employ immutable objects, because they worry about the cost of creating a new object as opposed to updating an object in place. The impact of object creation is often overestimated, and can be offset by some of the efficiencies associated with immutable objects. These include decreased overhead due to garbage collection, and the elimination of code needed to protect mutable objects from corruption.
The following subsections take a class whose instances are mutable and derives a class with immutable instances from it. In so doing, they give general rules for this kind of conversion and demonstrate some of the advantages of immutable objects.
Source
An immutable object is the one you cannot modify after you create it. A typical example are string literals.
A D programming language, which becomes increasingly popular, has a notion of "immutability" through "invariant" keyword. Check this Dr.Dobb's article about it - http://dobbscodetalk.com/index.php?option=com_myblog&show=Invariant-Strings.html&Itemid=29 . It explains the problem perfectly.
Related
Given a string
String x = "hello";
As far as I remember, x now is a constant with a reference but when in the middle of the code I change value to
x = "hey";
hello stays stuck in the memory and the GC frees it. Is it true ? If not, could you explain more how the GC work?
As far as I remember, x now is a constant
No. The reference is not a constant. It is a reference to an immutable, but the reference itself (the "pointer") is not (crediting Turing85 for the clarification). For x to be considered a constant, it needs to be declared with the keyword final (then the reference itself will be a constant). For example, final String x = "hello";. Because it is not a constant, it allows you to change the reference later on in the code x = "hey";. THAT SAID, because of how Java manages strings (and other immutables), "hello" and "hey" are indeed constants. They are "permanent" members of the String pool. Meaning that, their life cycle is the same as the life of your program. I think there are ways change this, but by default I am pretty sure this is the case. So, if you create multiple objects with the same String value:
String a = "hello"; // will add "hello" to String pool if it doesn't exist there
String b = "hello"; // finds "hello" in the string pool and returns back the same reference to this variable
String c = "hello"; // same as `b` above
However, if you create a new String using the copy constructor:
String d = new String("hello"); // Will add new "hello" to heap even though it already exist in string pool. (Thanks Holger!)
Explaining in detail how the GC works is not very easy to explain; especially in a single SO post. I don't even fully understand it myself. But, since the purpose of the pool is to eliminate the need to create "infinite" copies of the same thing over and over, "hello" and "hey" won't be garbage-collected. As for the variable x, it will be handled by the GC depending on many factors, one of which is scope. Also, there is no guarantee as to when (if ever) the GC will cleanup an unused resource.
I think this is accurate, albeit somewhat ambiguous or incomplete. But at a high-level, this is how GC will work in this case to the best of my knowledge.
UPDATE: Thanks to #Holger for bringing this to my attention. When a String Literal is assigned to a variable, it will first look for such value in the String Pool to see if such string exists. If it doesn't, it creates it and assigns it to the variable. HOWEVER, when a string is created with the new operator, it will add the new string JUST into Heap Memory. That's the main difference between assigning string literals vs. new strings to a variable.
String De-duplication:
Strings consume a lot of memory in any application.Whenever the garbage collector visits String objects it takes note of the char arrays. It takes their hash value and stores it alongside with a weak reference to the array. As soon as it finds another String which has the same hash code it compares them char by char.If they match as well, one String will be modified and point to the char array of the second String. The first char array then is no longer referenced anymore and can be garbage collected.
String Pool:
All strings used by the java program are stored here. If two variables are initialized to the same string value. Two strings are not created in the memory, there will be only one copy stored in memory and both will point to the same memory location.
So java already takes care of not creating duplicate strings in the heap by checking if the string exists in the string pool. Then what is the purpose of string de-duplication?
If there is a code as follows
String myString_1 = new String("Hello World");
String myString_2 = new String("Hello World");
two strings are created in memory even though they are same. I cannot think of any scenario other than this where string de-duplication is useful. Obviously I must be missing something. What I am I missing?
Thanks In Advance
The string pool applies only to strings added to it explicitly, or used as constants in the application. It does not apply to strings created dynamically during the lifetime of the application. String deduplication, however, applies to all strings.
Compile time vs run time
String pool refers to string constants that are known at compile time.
String deduplication would help you if you happen to retrieve (or construct) the same string a million times at run time, e.g. reading it from a file, a HTTP request or any other way.
String de-duplication enjoys the extra level of indirection built into String:
With a string pool, you are limited to returning the same object for two identical strings
String de-duplication lets you have multiple distinct String objects sharing the same content.
This translates into removing a limitation of de-duplicating on creation: your application could keep creating new String objects with identical content while using very little extra memory, because the content of the strings would be shared. This process can be done on a completely unrelated schedule - for example, in the background, while your application does not need much of the CPU resources. Since the identity of the String object does not change, de-duplication can be completely hidden form your application.
Just to add to the answers above, on older VM's the string pool is not garbage collected (this has changed now, but don't rely on that). It contains strings which are used as constants in the application, and so will always be needed. If you continually put all your strings in the string pool, you might quickly run out of memory. On top of that, de-duplication is a relatively expensive process, if you know you only need the string for a very short period of time, and you have enough memory.
For these reasons, strings are not put in the string pool automatically. You have to do it explicitly by calling string.intern().
From documentation :
"Initializes a newly created String object so that it represents
the same sequence of characters as the argument; in other words, the
newly created string is a copy of the argument string. Unless an
explicit copy of original is needed, use of this constructor is
unnecessary since Strings are immutable."
So my sense says, this constructor in String class is not needed normally like you have used above. I guess that constructor is provided merely for the sake of completeness or if you do not want to share that copy (kind of unnecessary now, refer here what I am talking about) but still other constructors are useful like getting an String object from char array and so on..
I cannot think of any scenario other than this where string de-duplication is useful.
Well one other (much more) frequent scenario is the use of StringBuilders. In the toString() method of the StringBuilder class, it clearly creates a new instance in memory:
public final class StringBuilder extends AbstractStringBuilder
implements java.io.Serializable, CharSequence
{
...
#Override
public String toString() {
// Create a copy, don't share the array
return new String(value, 0, count);
}
...
}
Same thing for its thread-safe version StringBuffer:
public final class StringBuffer extends AbstractStringBuilder
implements java.io.Serializable, CharSequence
{
...
#Override
public synchronized String toString() {
if (toStringCache == null) {
toStringCache = Arrays.copyOfRange(value, 0, count);
}
return new String(toStringCache, true);
}
...
}
In applications that rely heavily on this, string de-duplication may reduce memory usage.
I've been studying Java String for a while. The following questions are based on the below posts
Java String is special
Immutability of String in java
Immutability:
Now, going by the immutability, the String class has been designed so that the values in the common pool can be reused in other places/variables. This holds good if the String was created as
String a = "Hello World!";
However, if I create String like
String b = new String("Hello World!");
why is this immutable as well? (or is it?). Since this has a dedicated heap memory, I should be able to modify this without affecting any other variable. So by design, was there any other reason why String as a whole is considered immutable? Or is my above assumption wrong?
Second thing I wanted to ask was about the common string pool. If I create a string object as
String c = "";
is an empty entry created in the pool?
Is there any post already on these? If so, could someone share the link?
new String() is an expression that produces a String ... and a String is immutable, no matter how it is produced.
(Asking if new String() is mutable or not is nonsensical. It is program code, not a value. But I take it that that is not what you really meant.)
If I create a string object as String c = ""; is an empty entry created in the pool?
Yes; that is, an entry is created for the empty string. There is nothing special about an empty String.
(To be pedantic, the pool entry for "" gets created long before your code is executed. In fact, it is created when your code is loaded ... or possibly even earlier than that.)
So, I was wanted to know whether the new heap object is immutable as well, ...
Yes it is. But the immutability is a fundamental property of String objects. All String objects.
You see, the String API simply does not provide any methods for changing a String. So (apart from some dangerous and foolish1 tricks using reflection), you can't mutate a String.
and if so what was the purpose?.
The primary reason that Java String is designed as an immutable class is simplicity. It makes it easier to write correct programs, and read / reason about other people's code if the core string class provides an immutable interface.
An important second reason is that the immutability of String has fundamental implications for the Java security model. But I don't think this was a driver in the original language design ... in Java 1.0 and earlier.
Going by the answer, I gather that other references to the same variable is one of the reasons. Please let me know if I am right in understanding this.
No. It is more fundamental than that. Simply, all String objects are immutable. There is no complicated special case reasoning required to understand this. It just >>is<<.
For the record, if you want a mutable "string-like" object in Java, you can use StringBuilder or StringBuffer. But these are different types to String.
1 - The reason these tricks are (IMO) dangerous and foolish is that they affect the values of strings that are potentially shared by other parts of your application via the string pool. This can cause chaos ... in ways that the next guy maintaining your code has little chance of tracking down.
String is immutable irrespective of how it is instantiated
1) Short answer is yes, new String() is immutable too.
Because every possible mutable operation (like replace,toLowerCase etcetra) that you perform on String does not affect the original String instance and returns you a new instance.
You may check this in Javadoc for String. Each public method of String that is exposed returns a new String instance and does not alter the present instance on which you called the method.
This is very helpful in Multi-threaded environment as you don't have to think about mutability (someone will change the value) every time you pass or share the String around. String can easily be the most used data type, so the designers have blessed us all to not think about mutability everytime and saved us a lot of pain.
Immutability allowed String pool or caching
It is because of immutability property that the internal pool of string was possible, as when same String value is required at some other place then that immutable reference is returned. If String would have been mutable then it would not have been possible to share Strings like this to save memory.
String immutablity was not because of pooling, but immutability has more benefits attached to it.
String interning or pooling is an example of Flyweight Design pattern
2) Yes it will be interned like any other String as a blank String is also as much a String as other String instances.
References:
Immutability benefits of String
The Java libraries are heavily optimized around the constraint that any String object is immutable, regardless of how that object is constructed. Even if you create your b using new, other code that you pass that instance to will treat the value as immutable. This is an example of the Value Object pattern, and all of the advantages (thread-safety, no need to make private copies) apply.
The empty string "" is a legitimate String object just like anything else, it just happens to have no internal contents, and since all compile-time constant strings are interned, I'll virtually guarantee that some runtime library has already caused it to be added to the pool.
1) The immutable part isn't because of the pool; it just makes the pool possible in the first place. Strings are often passed as arguments to other functions or even shared with other threads; Making Strings immutable was a design decision to make reasoning in such situations easier. So yes - Strings in Java are always immutable, no matter how you create them (note that it's possible to have mutable strings in java - just not with the String class).
2) Yes. Probably. I'm not actually 100% sure, but that should be the case.
This isn't strictly an answer to your question, but if behind your question is a wish to have mutable strings that you can manipulate, you should check out the StringBuilder class, which implements many of the exact same methods that String has but also adds methods to change the current contents.
Once you've built your string in such a way that you're content with it, you simply call toString() on it in order to convert it to an ordinary String that you can pass to library routines and other functions that only take Strings.
Also, both StringBuilder and String implements the CharSequence interface, so if you want to write functions in your own code that can use both mutable and immutable strings, you can declare them to take any CharSequence object.
From the Java Oracle documentation:
Strings are constant; their values cannot be changed after they are
created.
And again:
String buffers support mutable strings. Because String
objects are immutable they can be shared.
Generally speaking: "all primitive" (or related) object are immutable (please, accept my lack of formalism).
Related post on Stack Overflow:
Immutability of Strings in Java
Is Java String immutable?
Is Java "pass-by-reference" or "pass-by-value"?
Is a Java string really immutable?
String is immutable. What exactly is the meaning?
what the reason behind making string immutable in java?
About the object pool: object pool is a java optimization which is NOT related to immutable as well.
String is immutable means that you cannot change the object itself no matter how you created it.And as for the second question: yes it will create an entry.
It's the other way around, actually.
[...] the String class has been designed so that the values in the common pool can be reused in other places/variables.
No, the String class is immutable so that you can safely reference an instance of it, without worrying about it being modified from another part of your program. This is why pooling is possible in the first place.
So, consider this:
// this string literal is interned and referenced by 'a'
String a = "Hello World!";
// creates a new instance by copying characters from 'a'
String b = new String(a);
Now, what happens if you simply create a reference to your newly created b variable?
// 'c' now points to the same instance as 'b'
String c = b;
Imagine that you pass c (or, more specifically, the object it is referencing) to a method on a different thread, and continue working with the same instance on your main thread. And now imagine what would happen if strings were mutable.
Why is this the case anyway?
If nothing else, it's because immutable objects make multithreading much simpler, and usually even faster. If you share a mutable object (that would be any stateful object, with mutable private/public fields or properties) between different threads, you need to take special care to ensure synchronized access (mutexes, semaphores). Even with this, you need special care to ensure atomicity in all your operations. Multithreading is hard.
Regarding the performance implications, note that quite often copying the entire string into a new instance, in order to change even a single character, is actually faster than inducing an expensive context switch due to synchronization constructs needed to ensure thread-safe access. And as you've mentioned, immutability also offers interning possibilities, which means it can actually help reduce memory usage.
It's generally a pretty good idea to make as much stuff immutable as your can.
1) Immutability: String will be immutable if you create it using new or other way for security reasons
2) Yes there will be a empty entry in the string pool.
You can better understand the concept using the code
String s1 = new String("Test");
String s2 = new String("Test");
String s3 = "Test";
String s4 = "Test";
System.out.println(s1==s2);//false
System.out.println(s1==s3);//false
System.out.println(s2==s3);//false
System.out.println(s4==s3);//true
Hope this will help your query. You can always check the source code for String class in case of better understanding Link : http://grepcode.com/file/repository.grepcode.com/java/root/jdk/openjdk/6-b14/java/lang/String.java
1-String is immutable.See this:
Is a Java string really immutable?
So you can create it with many way.
2-Short answer: Yes will be empty.
Strings created will always be immutable regardless of how they are created.
Answers to your questions:
The only difference is:
When string is created like -- {String a = "Hello World!";} then only one object gets created.
And when it is created like -- {String b = new String("Hello World!");} then two objects get created. The first one, because you have used 'new' keyword and the second one because of the String property.
Yes, for sure. There will be an empty entry created in the pool.
String is immutable because it does not provide you with a mean to modify it. It is design to avoid any tampering (it is final, the underlying array is not supposed to be touched ...).
Identically, Integer is immutable, because there is no way to modify it.
It does not matter how you create it.
Immutability is not a feature of new, it is a feature of the class String. It has no mutator methods, so it immutable.
String A = "Test"
String B = "Test"
Now String B called"Test".toUpperCase()which change the same object into"TEST", soAwill also be"TEST"` which is not desirable.
Note that in your example, the reference is changed and not the object it refers to, i.e. b as a reference may be changed and refer to a new object. But that new object is immutable, which means its contents will not be changed "after" calling the constructor.
You can change a string using b=b+"x"; or b=new String(b);, and the content of variable a seem to change, but don't confuse immutability of a reference (here variable b) and the object it is referring to (think of pointers in C). The object that the reference is pointing to will remain unchanged after its creation.
If you need to change a string by changing the contents of the object (instead of changing the reference), you can use StringBuffer, which is a mutable version of String.
We all know that String is immutable in Java, but check the following code:
String s1 = "Hello World";
String s2 = "Hello World";
String s3 = s1.substring(6);
System.out.println(s1); // Hello World
System.out.println(s2); // Hello World
System.out.println(s3); // World
Field field = String.class.getDeclaredField("value");
field.setAccessible(true);
char[] value = (char[])field.get(s1);
value[6] = 'J';
value[7] = 'a';
value[8] = 'v';
value[9] = 'a';
value[10] = '!';
System.out.println(s1); // Hello Java!
System.out.println(s2); // Hello Java!
System.out.println(s3); // World
Why does this program operate like this? And why is the value of s1 and s2 changed, but not s3?
String is immutable* but this only means you cannot change it using its public API.
What you are doing here is circumventing the normal API, using reflection. The same way, you can change the values of enums, change the lookup table used in Integer autoboxing etc.
Now, the reason s1 and s2 change value, is that they both refer to the same interned string. The compiler does this (as mentioned by other answers).
The reason s3 does not was actually a bit surprising to me, as I thought it would share the value array (it did in earlier version of Java, before Java 7u6). However, looking at the source code of String, we can see that the value character array for a substring is actually copied (using Arrays.copyOfRange(..)). This is why it goes unchanged.
You can install a SecurityManager, to avoid malicious code to do such things. But keep in mind that some libraries depend on using these kind of reflection tricks (typically ORM tools, AOP libraries etc).
*) I initially wrote that Strings aren't really immutable, just "effective immutable". This might be misleading in the current implementation of String, where the value array is indeed marked private final. It's still worth noting, though, that there is no way to declare an array in Java as immutable, so care must be taken not to expose it outside its class, even with the proper access modifiers.
As this topic seems overwhelmingly popular, here's some suggested further reading: Heinz Kabutz's Reflection Madness talk from JavaZone 2009, which covers a lot of the issues in the OP, along with other reflection... well... madness.
It covers why this is sometimes useful. And why, most of the time, you should avoid it. :-)
In Java, if two string primitive variables are initialized to the same literal, it assigns the same reference to both variables:
String Test1="Hello World";
String Test2="Hello World";
System.out.println(test1==test2); // true
That is the reason the comparison returns true. The third string is created using substring() which makes a new string instead of pointing to the same.
When you access a string using reflection, you get the actual pointer:
Field field = String.class.getDeclaredField("value");
field.setAccessible(true);
So change to this will change the string holding a pointer to it, but as s3 is created with a new string due to substring() it would not change.
You are using reflection to circumvent the immutability of String - it's a form of "attack".
There are lots of examples you can create like this (eg you can even instantiate a Void object too), but it doesn't mean that String is not "immutable".
There are use cases where this type of code may be used to your advantage and be "good coding", such as clearing passwords from memory at the earliest possible moment (before GC).
Depending on the security manager, you may not be able to execute your code.
You are using reflection to access the "implementation details" of string object. Immutability is the feature of the public interface of an object.
Visibility modifiers and final (i.e. immutability) are not a measurement against malicious code in Java; they are merely tools to protect against mistakes and to make the code more maintainable (one of the big selling points of the system). That is why you can access internal implementation details like the backing char array for Strings via reflection.
The second effect you see is that all Strings change while it looks like you only change s1. It is a certain property of Java String literals that they are automatically interned, i.e. cached. Two String literals with the same value will actually be the same object. When you create a String with new it will not be interned automatically and you will not see this effect.
#substring until recently (Java 7u6) worked in a similar way, which would have explained the behaviour in the original version of your question. It didn't create a new backing char array but reused the one from the original String; it just created a new String object that used an offset and a length to present only a part of that array. This generally worked as Strings are immutable - unless you circumvent that. This property of #substring also meant that the whole original String couldn't be garbage collected when a shorter substring created from it still existed.
As of current Java and your current version of the question there is no strange behaviour of #substring.
String immutability is from the interface perspective. You are using reflection to bypass the interface and directly modify the internals of the String instances.
s1 and s2 are both changed because they are both assigned to the same "intern" String instance. You can find out a bit more about that part from this article about string equality and interning. You might be surprised to find out that in your sample code, s1 == s2 returns true!
Which version of Java are you using? From Java 1.7.0_06, Oracle has changed the internal representation of String, especially the substring.
Quoting from Oracle Tunes Java's Internal String Representation:
In the new paradigm, the String offset and count fields have been removed, so substrings no longer share the underlying char [] value.
With this change, it may happen without reflection (???).
There are really two questions here:
Are strings really immutable?
Why is s3 not changed?
To point 1: Except for ROM there is no immutable memory in your computer. Nowadays even ROM is sometimes writable. There is always some code somewhere (whether it's the kernel or native code sidestepping your managed environment) that can write to your memory address. So, in "reality", no they are not absolutely immutable.
To point 2: This is because substring is probably allocating a new string instance, which is likely copying the array. It is possible to implement substring in such a way that it won't do a copy, but that doesn't mean it does. There are tradeoffs involved.
For example, should holding a reference to reallyLargeString.substring(reallyLargeString.length - 2) cause a large amount of memory to be held alive, or only a few bytes?
That depends on how substring is implemented. A deep copy will keep less memory alive, but it will run slightly slower. A shallow copy will keep more memory alive, but it will be faster. Using a deep copy can also reduce heap fragmentation, as the string object and its buffer can be allocated in one block, as opposed to 2 separate heap allocations.
In any case, it looks like your JVM chose to use deep copies for substring calls.
To add to the #haraldK's answer - this is a security hack which could lead to a serious impact in the app.
First thing is a modification to a constant string stored in a String Pool. When string is declared as a String s = "Hello World";, it's being places into a special object pool for further potential reusing. The issue is that compiler will place a reference to the modified version at compile time and once the user modifies the string stored in this pool at runtime, all references in code will point to the modified version. This would result into a following bug:
System.out.println("Hello World");
Will print:
Hello Java!
There was another issue I experienced when I was implementing a heavy computation over such risky strings. There was a bug which happened in like 1 out of 1000000 times during the computation which made the result undeterministic. I was able to find the problem by switching off the JIT - I was always getting the same result with JIT turned off. My guess is that the reason was this String security hack which broke some of the JIT optimization contracts.
According to the concept of pooling, all the String variables containing the same value will point to the same memory address. Therefore s1 and s2, both containing the same value of “Hello World”, will point towards the same memory location (say M1).
On the other hand, s3 contains “World”, hence it will point to a different memory allocation (say M2).
So now what's happening is that the value of S1 is being changed (by using the char [ ] value). So the value at the memory location M1 pointed both by s1 and s2 has been changed.
Hence as a result, memory location M1 has been modified which causes change in the value of s1 and s2.
But the value of location M2 remains unaltered, hence s3 contains the same original value.
The reason s3 does not actually change is because in Java when you do a substring the value character array for a substring is internally copied (using Arrays.copyOfRange()).
s1 and s2 are the same because in Java they both refer to the same interned string. It's by design in Java.
String is immutable, but through reflection you're allowed to change the String class. You've just redefined the String class as mutable in real-time. You could redefine methods to be public or private or static if you wanted.
Strings are created in permanent area of the JVM heap memory. So yes, it's really immutable and cannot be changed after being created.
Because in the JVM, there are three types of heap memory:
1. Young generation
2. Old generation
3. Permanent generation.
When any object are created, it goes into the young generation heap area and PermGen area reserved for String pooling.
Here is more detail you can go and grab more information from:
How Garbage Collection works in Java .
[Disclaimer this is a deliberately opinionated style of answer as I feel a more "don't do this at home kids" answer is warranted]
The sin is the line field.setAccessible(true); which says to violate the public api by allowing access to a private field. Thats a giant security hole which can be locked down by configuring a security manager.
The phenomenon in the question are implementation details which you would never see when not using that dangerous line of code to violate the access modifiers via reflection. Clearly two (normally) immutable strings can share the same char array. Whether a substring shares the same array depends on whether it can and whether the developer thought to share it. Normally these are invisible implementation details which you should not have to know unless you shoot the access modifier through the head with that line of code.
It is simply not a good idea to rely upon such details which cannot be experienced without violating the access modifiers using reflection. The owner of that class only supports the normal public API and is free to make implementation changes in the future.
Having said all that the line of code is really very useful when you have a gun held you your head forcing you to do such dangerous things. Using that back door is usually a code smell that you need to upgrade to better library code where you don't have to sin. Another common use of that dangerous line of code is to write a "voodoo framework" (orm, injection container, ...). Many folks get religious about such frameworks (both for and against them) so I will avoid inviting a flame war by saying nothing other than the vast majority of programmers don't have to go there.
String is immutable in nature Because there is no method to modify String object.
That is the reason They introduced StringBuilder and StringBuffer classes
This is a quick guide to everything
// Character array
char[] chr = {'O', 'K', '!'};
// this is String class
String str1 = new String(chr);
// this is concat
str1 = str1.concat("another string's ");
// this is format
System.out.println(String.format(str1 + " %s ", "string"));
// this is equals
System.out.println(str1.equals("another string"));
//this is split
for(String s: str1.split(" ")){
System.out.println(s);
}
// this is length
System.out.println(str1.length());
//gives an score of the total change in the length
System.out.println(str1.compareTo("OK!another string string's"));
// trim
System.out.println(str1.trim());
// intern
System.out.println(str1.intern());
// character at
System.out.println(str1.charAt(5));
// substring
System.out.println(str1.substring(5, 12));
// to uppercase
System.out.println(str1.toUpperCase());
// to lowerCase
System.out.println(str1.toLowerCase());
// replace
System.out.println(str1.replace("another", "hello"));
// output
// OK!another string's string
// false
// OK!another
// string's
// 20
// 7
// OK!another string's
// OK!another string's
// o
// other s
// OK!ANOTHER STRING'S
// ok!another string's
// OK!hello string's
I actually thought that I had a good idea of how passing values in Java actually work, since that was part of the SCJP cert which I have passed. That was until today, when I at work discovered a method like this:
public void toCommand(Stringbuffer buf) {
buf.append("blablabla");
}
Then the caller of that method used the function like this:
StringBuffer buf = new StringBuffer();
toCommand(buf);
String str = buf.toString();
Now I thought that that code would give str the value "", but it actually give it the value from the mehod. How is this possible? I thought things didnt work like that in Java?
Either way... it should be considered a bad practice to write code like this in Java, right? Because I can imagine it can bring some confusion with it.
I actually spent some time searching on this, but my interpretation of what these sources are saying, is that it shouldnt work. What am I missing?
http://www.yoda.arachsys.com/java/passing.html
http://javadude.com/articles/passbyvalue.htm
Sebastian
Java is pass-by-value. The value of an object reference is an object reference, not the object itself. And so the toCommand method receives a copy of the value, which is a reference to the object — the same object that the caller is referencing.
This is exactly the same as when you're referencing an object from two variables:
StringBuffer buf1;
StringBuffer buf2;
buf1 = new StringBuffer();
buf2 = buf1; // Still ONE object; there are two references to it
buf1.append("Hi there");
System.out.println(buf2.toString()); // "Hi there"
Gratuitous ASCII art:
+--------------------+
buf1--------->| |
| === The Object === |
| |
buf2--------->| Data: |
| * foo = "bar" |
| * x = 27 |
| |
+--------------------+
Another way to think of it is that the JVM has a master list of all objects, indexed by an ID. We create an object (buf1 = new StringBuffer();) and the JVM assigns the object the ID 42 and stores that ID in buf1 for us. Whenever we use buf1, the JVM gets the value 42 from it and looks up the object in its master list, and uses the object. When we do buf2 = buf1;, the variable buf2 gets a copy of the value 42, and so when we use buf2, the JVM sees object reference #42 and uses that same object. This is not a literal explanation (though from a stratospheric viewpoint, and if you read "JVM" as "JVM and memory manager and OS", it's not a million miles off), but helpful for thinking about what object references actually are.
With that background, you can see how toCommand gets a reference (42 or whatever), not the actual StringBuffer object data. And so operations on it look it up in the master list and alter its state (since it holds state information and allows us to change it). The caller sees the changes to the object's state because the object holds the state, the reference just points to the object.
Either way... it should be considered a bad practice to write code like this in Java, right?
Not at all, it's normal practice. It would be very hard to use Java (or most other OOP languages) without doing this. Objects are big compared to primitives like int and long, and so they're expensive to move around; object references are the size of primitives, so they're easily passed around. Also, having copies of things makes it difficult for various parts of a system to interact. Having references to shared objects makes it quite easy.
StringBuffer is mutable. The toCommand() method gets a reference to that objects which is passed by value (the reference), the reference allows the method to change the mutable StringBuffer.
If you are thinking about why we cannot do this with String it is because String is immutable in which case it will result in the creation of another String object and not having the changes reflected in the object of which the reference is passed.
And I don't see why it should be bad practice.
This is because the in the method, when you pass an object, a copy of a reference to the object is passed by value. Consider the example below:
public class Test {
public static void modifyBuff(StringBuffer b) {
b.append("foo");
}
public static void tryToNullifyBuff(StringBuffer b) {
b = null; // this will not affect the original reference since
// the once passed (by value) is a copy
}
public static void main(String[] args) {
StringBuffer buff = new StringBuffer(); // buff is a reference
// to StringBuffer object
modifyBuff(buff);
System.out.println(buff); // will print "foo"
tryToNullifyBuff(buff); // this has no effect on the original reference 'buff'
System.out.println(buff); // will still print "foo" because a copy of
// reference buff is passed to tryToNullifyBuff()
// which is made to reference null
// inside the method leaving the 'buff' reference intact
}
}
This can be done with other mutable objects like Collection classes for example. And this is
not at all a bad practice, in fact certain designs actively use this pattern.
str has the value "blablabla" because a reference to the StringBuilder instance is passed to toCommand().
There is only one StringBuffer instance created here - and you pass a reference to it to the toCommand() method. Therefore any methods invoked on the StringBuffer in the toCommand() method are invoked on the same instance of the StringBuffer in the calling method.
It is not always bad practise. Consider the task to assemble an email, then you could do something like:
StringBuilder emailBuilder = new StringBuilder();
createHeader(emailBuilder);
createBody(emailBuilder);
createFooter(emailBuilder);
sendEmail(emailBuilder.toString());
It surely could be used to create confusion and for public API one should add a note or two to the javadoc, if the value at the passed reference is changed.
Another prominent example from the Java API:
Collections.sort(list);
And as other already explained and just to complete the answer: The value of the reference of the StringBuffer (<-- start using StringBuilder instead!) is passed to toCommand so outside and inside the toCommand method you access the same StringBuffer instance.
Since you get a reference to the instance you can call all the methods on it, but you can't assign it to anything else.
public void toCommand(Stringbuffer buf) {
buf.append("blablabla"); // okay
buf = new Stringbuffer(); // no "effect" outside this method
}
When you pass an object to a method, unlike in C++, only the reference to the object gets copied. So when a mutable object is passed to the method, you can change it and the change gets reflected in the calling method.
In case you're too much into C/C++ programming, you should know that in java, pass by reference(in C++ lingo) is the default(and the only) way of passing arguments to methods.
Just in response to some of the comments about the SCJP (sorry don't have permissions to leave comments - apologies if leaving an answer is not the right way to do this).
In defence of the SCJP exam, passing object references as method parameters AND the differences between immutable String objects and StringBuffer / StringBuilder objects are part of the exam - see sections 3.1 and 7.3 here:
http://www.javadeveloper.co.in/scjp/scjp-exam-objectives.html
Both topics are covered quite extensively in the Kathy Sierra / Bert Bates study guide to the exam (which is the de facto official study guide to the exam).