Consider the following example.
String str = new String();
str = "Hello";
System.out.println(str); //Prints Hello
str = "Help!";
System.out.println(str); //Prints Help!
Now, in Java, String objects are immutable. Then how come the object str can be assigned value "Help!". Isn't this contradicting the immutability of strings in Java? Can anybody please explain me the exact concept of immutability?
Edit:
Ok. I am now getting it, but just one follow-up question. What about the following code:
String str = "Mississippi";
System.out.println(str); // prints Mississippi
str = str.replace("i", "!");
System.out.println(str); // prints M!ss!ss!pp!
Does this mean that two objects are created again ("Mississippi" and "M!ss!ss!pp!") and the reference str points to a different object after replace() method?
str is not an object, it's a reference to an object. "Hello" and "Help!" are two distinct String objects. Thus, str points to a string. You can change what it points to, but not that which it points at.
Take this code, for example:
String s1 = "Hello";
String s2 = s1;
// s1 and s2 now point at the same string - "Hello"
Now, there is nothing1 we could do to s1 that would affect the value of s2. They refer to the same object - the string "Hello" - but that object is immutable and thus cannot be altered.
If we do something like this:
s1 = "Help!";
System.out.println(s2); // still prints "Hello"
Here we see the difference between mutating an object, and changing a reference. s2 still points to the same object as we initially set s1 to point to. Setting s1 to "Help!" only changes the reference, while the String object it originally referred to remains unchanged.
If strings were mutable, we could do something like this:
String s1 = "Hello";
String s2 = s1;
s1.setCharAt(1, 'a'); // Fictional method that sets character at a given pos in string
System.out.println(s2); // Prints "Hallo"
Edit to respond to OP's edit:
If you look at the source code for String.replace(char,char) (also available in src.zip in your JDK installation directory -- a pro tip is to look there whenever you wonder how something really works) you can see that what it does is the following:
If there is one or more occurrences of oldChar in the current string, make a copy of the current string where all occurrences of oldChar are replaced with newChar.
If the oldChar is not present in the current string, return the current string.
So yes, "Mississippi".replace('i', '!') creates a new String object. Again, the following holds:
String s1 = "Mississippi";
String s2 = s1;
s1 = s1.replace('i', '!');
System.out.println(s1); // Prints "M!ss!ss!pp!"
System.out.println(s2); // Prints "Mississippi"
System.out.println(s1 == s2); // Prints "false" as s1 and s2 are two different objects
Your homework for now is to see what the above code does if you change s1 = s1.replace('i', '!'); to s1 = s1.replace('Q', '!'); :)
1 Actually, it is possible to mutate strings (and other immutable objects). It requires reflection and is very, very dangerous and should never ever be used unless you're actually interested in destroying the program.
The object that str references can change, but the actual String objects themselves cannot.
The String objects containing the string "Hello" and "Help!" cannot change their values, hence they are immutable.
The immutability of String objects does not mean that the references pointing to the object cannot change.
One way that one can prevent the str reference from changing is to declare it as final:
final String STR = "Hello";
Now, trying to assign another String to STR will cause a compile error.
Light_handle I recommend you take a read of Cup Size -- a story about variables and Pass-by-Value Please (Cup Size continued). This will help a lot when reading the posts above.
Have you read them? Yes. Good.
String str = new String();
This creates a new "remote control" called "str" and sets that to the value new String() (or "").
e.g. in memory this creates:
str --- > ""
str = "Hello";
This then changes the remote control "str" but does not modify the original string "".
e.g. in memory this creates:
str -+ ""
+-> "Hello"
str = "Help!";
This then changes the remote control "str" but does not modify the original string "" or the object that the remote control currently points to.
e.g. in memory this creates:
str -+ ""
| "Hello"
+-> "Help!"
Lets break it into some parts
String s1 = "hello";
This Statement creates string containing hello and occupy space in memory i.e. in Constant String Pool and and assigned it to reference object s1
String s2 = s1;
This statement assigns the same string hello to new reference s2
__________
| |
s1 ---->| hello |<----- s2
|__________|
Both references are pointing to the same string so output the same value as follows.
out.println(s1); // o/p: hello
out.println(s2); // o/p: hello
Though String is immutable, assignment can be possible so the s1 will now refer to new value stack.
s1 = "stack";
__________
| |
s1 ---->| stack |
|__________|
But what about s2 object which is pointing to hello it will be as it is.
__________
| |
s2 ---->| hello |
|__________|
out.println(s1); // o/p: stack
out.println(s2); // o/p: hello
Since String is immutable Java Virtual Machine won't allow us to modify string s1 by its method. It will create all new String object in pool as follows.
s1.concat(" overflow");
___________________
| |
s1.concat ----> | stack overflow |
|___________________|
out.println(s1); // o/p: stack
out.println(s2); // o/p: hello
out.println(s1.concat); // o/p: stack overflow
Note if String would be mutable then the output would have been
out.println(s1); // o/p: stack overflow
Now you might be surprised why String has such methods like concat() to modify. Following snippet will clear your confusion.
s1 = s1.concat(" overflow");
Here we are assigning modified value of string back to s1 reference.
___________________
| |
s1 ---->| stack overflow |
|___________________|
out.println(s1); // o/p: stack overflow
out.println(s2); // o/p: hello
That's why Java decided String to be a final class Otherwise anyone can modify and change the value of string.
Hope this will help little bit.
The string object that was first referenced by str was not altered, all that you did was make str refer to a new string object.
The String will not change, the reference to it will. You are confusing immutability with the concept of final fields. If a field is declared as final, once it has been assigned, it cannot be reassigned.
Regarding the replace part of your question, try this:
String str = "Mississippi";
System.out.println(str); //Prints Mississippi
String other = str.replace("i", "!");
System.out.println(str); //still prints Mississippi
System.out.println(other); // prints M!ss!ss!pp!
Though java tries to ignore it, str is nothing more than a pointer. This means that when you first write str = "Hello";, you create an object that str points to. When you reassign str by writing str = "Help!";, a new object is created and the old "Hello" object gets garbage collected whenever java feels like it.
Immutability implies that the value of an instantiated object cannot change, you can never turn "Hello" into "Help!".
The variable str is a reference to an object, when you assign a new value to str you aren't changing the value of the object it references, you are referencing a different object.
String class is immutable, and you can not change value of immutable object.
But in case of String, if you change the value of string than it will create new string in string pool and than your string reference to that value not the older one. so by this way string is immutable.
Lets take your example,
String str = "Mississippi";
System.out.println(str); // prints Mississippi
it will create one string "Mississippi" and will add it to String pool
so now str is pointing to Mississippi.
str = str.replace("i", "!");
System.out.println(str); // prints M!ss!ss!pp!
But after above operation,
one another string will be created "M!ss!ss!pp!"
and it will be add to String pool. and
now str is pointing to M!ss!ss!pp!, not Mississippi.
so by this way when you will alter value of string object it will create one more object and will add it to string pool.
Lets have one more example
String s1 = "Hello";
String s2 = "World";
String s = s1 + s2;
this above three line will add three objects of string to string pool.
1) Hello
2) World
3) HelloWorld
For those wondering how to break String immutability in Java...
Code
import java.lang.reflect.Field;
public class StringImmutability {
public static void main(String[] args) {
String str1 = "I am immutable";
String str2 = str1;
try {
Class str1Class = str1.getClass();
Field str1Field = str1Class.getDeclaredField("value");
str1Field.setAccessible(true);
char[] valueChars = (char[]) str1Field.get(str1);
valueChars[5] = ' ';
valueChars[6] = ' ';
System.out.println(str1 == str2);
System.out.println(str1);
System.out.println(str2);
} catch (NoSuchFieldException e) {
e.printStackTrace();
} catch (SecurityException e) {
e.printStackTrace();
} catch (IllegalArgumentException e) {
e.printStackTrace();
} catch (IllegalAccessException e) {
e.printStackTrace();
}
}
}
Output
true
I am mutable
I am mutable
Use:
String s = new String("New String");
s.concat(" Added String");
System.out.println("String reference -----> "+s); // Output: String reference -----> New String
If you see here I use the concat method to change the original string, that is, "New String" with a string " Added String", but still I got the output as previous, hence it proves that you can not change the reference of object of String class, but if you do this thing by StringBuilder class it will work. It is listed below.
StringBuilder sb = new StringBuilder("New String");
sb.append(" Added String");
System.out.println("StringBuilder reference -----> "+sb);// Output: StringBuilder reference -----> New String Added String
Like Linus Tolvards said:
Talk is cheap. Show me the code
Take a look at this:
public class Test{
public static void main(String[] args){
String a = "Mississippi";
String b = "Mississippi";//String immutable property (same chars sequence), then same object
String c = a.replace('i','I').replace('I','i');//This method creates a new String, then new object
String d = b.replace('i','I').replace('I','i');//At this moment we have 3 String objects, a/b, c and d
String e = a.replace('i','i');//If the arguments are the same, the object is not affected, then returns same object
System.out.println( "a==b? " + (a==b) ); // Prints true, they are pointing to the same String object
System.out.println( "a: " + a );
System.out.println( "b: " + b );
System.out.println( "c==d? " + (c==d) ); // Prints false, a new object was created on each one
System.out.println( "c: " + c ); // Even the sequence of chars are the same, the object is different
System.out.println( "d: " + d );
System.out.println( "a==e? " + (a==e) ); // Same object, immutable property
}
}
The output is
a==b? true
a: Mississippi
b: Mississippi
c==d? false
c: Mississippi
d: Mississippi
a==e? true
So, remember two things:
Strings are immutable until you apply a method that manipulates and creates a new one (c & d cases).
Replace method returns the same String object if both parameters are the same
String is immutable. Which means that we can only change the reference.
String a = "a";
System.out.println("String a is referencing to "+a); // Output: a
a.concat("b");
System.out.println("String a is referencing to "+a); // Output: a
a = a.concat("b");
System.out.println("String a has created a new reference and is now referencing to "+a); // Output: ab
In Java, objects are generally accessed by references. In your piece of code str is a reference which is first assigned to "Hello" (an automatic created object or fetched from constant pool) and then you assigned another object "Help!" to same reference. A point to note is the reference is the same and modified, but objects are different. One more thing in your code you accessed three objects,
When you called new String().
When you assigned "hello".
When you assigned "help!".
Calling new String() creates a new object even if it exists in string pool, so generally it should not be used. To put a string created from new String () into string pool you can try the intern() method.
I hope this helps.
Immutability I can say is that you cannot change the String itself. Suppose you have String x, the value of which is "abc". Now you cannot change the String, that is, you cannot change any character/s in "abc".
If you have to change any character/s in the String, you can use a character array and mutate it or use StringBuilder.
String x = "abc";
x = "pot";
x = x + "hj";
x = x.substring(3);
System.out.println(x);
char x1[] = x.toCharArray();
x1[0] = 's';
String y = new String(x1);
System.out.println(y);
Output:
hj
sj
Or you can try:
public class Tester
{
public static void main(String[] args)
{
String str = "Mississippi";
System.out.println(str); // prints Mississippi
System.out.println(str.hashCode());
str = str.replace("i", "!");
System.out.println(str); // prints M!ss!ss!pp!
System.out.println(str.hashCode());
}
}
This will show how the hashcode changes.
String is immutable means that you cannot change the object itself, but you can change the reference to the object. When you called a = "ty", you are actually changing the reference of a to a new object created by the String literal "ty". Changing an object means to use its methods to change one of its fields (or the fields are public and not final, so that they can be updated from outside without accessing them via methods), for example:
Foo x = new Foo("the field");
x.setField("a new field");
System.out.println(x.getField()); // prints "a new field"
While in an immutable class (declared as final, to prevent modification via inheritance)(its methods cannot modify its fields, and also the fields are always private and recommended to be final), for example String, you cannot change the current String but you can return a new String, i.e:
String s = "some text";
s.substring(0,4);
System.out.println(s); // still printing "some text"
String a = s.substring(0,4);
System.out.println(a); // prints "some"
Here immutability means that instance can point to other reference but the original content of the string would not be modified at the original reference.
Let me explain by first example given by you.
First str is pointing to "Hello" ,its Ok upto this.
Second time its pointing to "Help!".
Here str started pointing to "Help!" and the reference of "Hello" string is lost and we can not get that back.
In fact when str would try to modify the existing content,then another new string will be generated and str will start to point at that reference.
So we see that string at original reference is not modified but that is safe at its reference and instance of object started pointing at different reference so immutability is conserve.
Super late to the answer, but wanted to put a concise message from author of the String class in Java
Strings are constant; their values cannot be changed after they are
created. String buffers support mutable strings. Because String
objects are immutable they can be shared.
It can be derived from this documentation that anything that changes string, returns different object (which could be new or interned and old).
The not so subtle hint about this should come from the function signature.
Think about it, 'Why did they make a function on an object return an object instead of status?'.
public String replace(char oldChar, char newChar)
Also one more source which makes this behaviour explicit (From replace function documentation)
Returns a new string resulting from replacing all occurrences of
oldChar in this string with newChar.
Source: https://docs.oracle.com/javase/7/docs/api/java/lang/String.html#replace(char,%20char)
author Lee Boynton
author Arthur van Hoff
author Martin Buchholz
author Ulf Zibis
Source: JavaDoc of String.
The Object string - methods itself is made to be "immutable".
This action produces no changes: "letters.replace("bbb", "aaa");"
But assigning data does cause changes to the Strings content to change:
letters = "aaa";
letters=null;
System.out.println(letters);
System.out.println(oB.hashCode());
System.out.println(letters);
letters = "bbbaaa";
System.out.println(oB.hashCode());
System.out.println(letters);
//The hashcode of the string Object doesn't change.
If HELLO is your String then you can't change HELLO to HILLO. This property is called immutability property.
You can have multiple pointer String variable to point HELLO String.
But if HELLO is char Array then you can change HELLO to HILLO. Eg,
char[] charArr = 'HELLO';
char[1] = 'I'; //you can do this
Programming languages have immutable data variables so that it can be used as keys in key, value pair.
I would explain it with simple example
consider any character array : e.g. char a[]={'h','e','l','l','o'};
and a string :
String s="hello";
on character array we can perform operations like printing only last three letters using iterating the array;
but in string we have to make new String object and copy required substring and its address will be in new string object.
e.g.
***String s="hello";
String s2=s.substrig(0,3);***
so s2 will have "hel";
String in Java in Immutable and Final just mean it can't be changed or modified:
Case 1:
class TestClass{
public static void main(String args[]){
String str = "ABC";
str.concat("DEF");
System.out.println(str);
}
}
Output: ABC
Reason: The object reference str is not changed in fact a new object
"DEF" is created which is in the pool and have no reference at all
(i.e lost).
Case 2:
class TestClass{
public static void main(String args[]){
String str="ABC";
str=str.concat("DEF");
System.out.println(str);
}
}
Output: ABCDEF
Reason: In this case str is now referring to a new object "ABCDEF"
hence it prints ABCDEF i.e. previous str object "ABC" is lost in pool with no reference.
Because String is immutable so changes will not occur if you will not assign the returned value of function to the string.so in your question assign value of swap function returned value to s.
s=swap(s, n1, n2) ;then the value of string s will change.
I was also getting the unchanged value when i was writing the program to get some permutations string(Although it is not giving all the permutations but this is for example to answer your question)
Here is a example.
> import java.io.*; public class MyString { public static void
> main(String []args)throws IOException { BufferedReader br=new
> BufferedReader(new InputStreamReader(System.in)); String
> s=br.readLine().trim(); int n=0;int k=0; while(n!=s.length()) {
> while(k<n){ swap(s,k,n); System.out.println(s); swap(s,k,n); k++; }
> n++; } } public static void swap(String s,int n1,int n2) { char temp;
> temp=s.charAt(n1); StringBuilder sb=new StringBuilder(s);
> sb.setCharAt(n1,s.charAt(n2)); sb.setCharAt(n2,temp); s=sb.toString();
> } }
but i was not getting the permuted values of the string from above code.So I assigned the returned value of the swap function to the string and got changed values of string. after assigning the returned value i got the permuted values of string.
/import java.util.*; import java.io.*; public class MyString { public static void main(String []args)throws IOException{
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
String s=br.readLine().trim(); int n=0;int k=0;
while(n!=s.length()){ while(k<n){ s=swap(s,k,n);
System.out.println(s); s=swap(s,k,n); k++; } n++; } }
public static String swap(String s,int n1,int n2){
char temp; temp=s.charAt(n1); StringBuilder sb=new StringBuilder(s); sb.setCharAt(n1,s.charAt(n2)); sb.setCharAt(n2,temp); s=sb.toString(); return s; } }
public final class String_Test {
String name;
List<String> list=new ArrayList<String>();
public static void main(String[] args) {
String_Test obj=new String_Test();
obj.list.add("item");//List will point to a memory unit- i.e will have one Hashcode value #1234
List<String> list2=obj.list; //lis1 also will point to same #1234
obj.list.add("new item");//Hashcode of list is not altered- List is mutable, so reference remains same, only value in that memory location changes
String name2=obj.name="Myname"; // name2 and name will point to same instance of string -Hashcode #5678
obj.name = "second name";// String is Immutable- New String HAI is created and name will point to this new instance- bcoz of this Hashcode changes here #0089
System.out.println(obj.list.hashCode());
System.out.println(list2.hashCode());
System.out.println(list3.hashCode());
System.out.println("===========");
System.out.println(obj.name.hashCode());
System.out.println(name2.hashCode());
}
}
Will produce out put something like this
1419358369
1419358369
103056
65078777
Purpose of Immutable object is that its value should not be altered once assigned.
It will return new object everytime you try to alter it based on the implementation.
Note: Stringbuffer instead of string can be used to avoid this.
To your last question :: u will have one reference , and 2 strings in string pool..
Except the reference will point to m!ss!ss!pp!
As per my knowledge,
a mutable string can be changed, and
an immutable string cannot be changed.
Here I want to change the value of String like this,
String str="Good";
str=str+" Morning";
and other way is,
StringBuffer str= new StringBuffer("Good");
str.append(" Morning");
In both the cases I am trying to alter the value of str. Can anyone tell me, what is difference in both case and give me clear picture of mutable and immutable objects.
Case 1:
String str = "Good";
str = str + " Morning";
In the above code you create 3 String Objects.
"Good" it goes into the String Pool.
" Morning" it goes into the String Pool as well.
"Good Morning" created by concatenating "Good" and " Morning". This guy goes on the Heap.
Note: Strings are always immutable. There is no, such thing as a mutable String. str is just a reference which eventually points to "Good Morning". You are actually, not working on 1 object. you have 3 distinct String Objects.
Case 2:
StringBuffer str = new StringBuffer("Good");
str.append(" Morning");
StringBuffer contains an array of characters. It is not same as a String.
The above code adds characters to the existing array. Effectively, StringBuffer is mutable, its String representation isn't.
What is difference between mutable and immutable String in java
immutable exist, mutable don't.
In Java, all strings are immutable. When you are trying to modify a String, what you are really doing is creating a new one. However, when you use a StringBuilder, you are actually modifying the contents, instead of creating a new one.
Java Strings are immutable.
In your first example, you are changing the reference to the String, thus assigning it the value of two other Strings combined: str + " Morning".
On the contrary, a StringBuilder or StringBuffer can be modified through its methods.
Mutable :- Changeable
Immutable :- Unchangeable
String in Java is immutable. However what does it mean to be mutable in programming context is the first question. Consider following class,
public class Dimension {
private int height;
private int width;
public Dimenstion() {
}
public void setSize(int height, int width) {
this.height = height;
this.width = width;
}
public getHeight() {
return height;
}
public getWidth() {
return width;
}
}
Now after creating the instance of Dimension we can always update it's attributes. Note that if any of the attribute, in other sense state, can be updated for instance of the class then it is said to be mutable. We can always do following,
Dimension d = new Dimension();
d.setSize(10, 20);// Dimension changed
d.setSize(10, 200);// Dimension changed
d.setSize(100, 200);// Dimension changed
Let's see in different ways we can create a String in Java.
String str1 = "Hey!";
String str2 = "Jack";
String str3 = new String("Hey Jack!");
String str4 = new String(new char[] {'H', 'e', 'y', '!'});
String str5 = str1 + str2;
str1 = "Hi !";
// ...
So,
str1 and str2 are String literals which gets created in String constant pool
str3, str4 and str5 are String Objects which are placed in Heap memory
str1 = "Hi!"; creates "Hi!" in String constant pool and it's totally different reference than "Hey!" which str1 referencing earlier.
Here we are creating the String literal or String Object. Both are different, I would suggest you to read following post to understand more about it.
Difference between string object and string literal
In any String declaration, one thing is common, that it does not modify but it gets created or shifted to other.
String str = "Good"; // Create the String literal in String pool
str = str + " Morning"; // Create String with concatenation of str + "Morning"
|_____________________|
|- Step 1 : Concatenate "Good" and " Morning" with StringBuilder
|- Step 2 : assign reference of created "Good Morning" String Object to str
How String became immutable ?
It's non changing behaviour, means, the value once assigned can not be updated in any other way. String class internally holds data in character array. Moreover, class is created to be immutable. Take a look at this strategy for defining immutable class.
Shifting the reference does not mean you changed it's value. It would be mutable if you can update the character array which is behind the scene in String class. But in reality that array will be initialized once and throughout the program it remains the same.
Why StringBuffer is mutable ?
As you already guessed, StringBuffer class is mutable itself as you can update it's state directly. Similar to String it also holds value in character array and you can manipulate that array by different methods i.e. append, delete, insert etc. which directly changes the character value array.
When you say str, you should be careful what you mean:
do you mean the variable str?
or do you mean the object referenced by str?
In your StringBuffer example you are not altering the value of str, and in your String example you are not altering the state of the String object.
The most poignant way to experience the difference would be something like this:
static void change(String in) {
in = in + " changed";
}
static void change(StringBuffer in) {
in.append(" changed");
}
public static void main(String[] args) {
StringBuffer sb = new StringBuffer("value");
String str = "value";
change(sb);
change(str);
System.out.println("StringBuffer: "+sb);
System.out.println("String: "+str);
}
In Java, all strings are immutable(Can't change). When you are trying to modify a String, what you are really doing is creating a new one.
Following ways we can create the string object
Using String literal
String str="java";
Using new keyword
String str = new String("java");
Using character array
char[] helloArray = { 'h', 'e', 'l', 'l', 'o', '.' };
String helloString = new String(helloArray);
coming to String immutability, simply means unmodifiable or unchangeable
Let's take one example
I'm initializing the value to the String literal s
String s="kumar";
Below I'm going to display the decimal representation of the location address using hashcode()
System.out.println(s.hashCode());
Simply printing the value of a String s
System.out.println("value "+s);
Okay, this time I'm inittializing value "kumar" to s1
String s1="kumar"; // what you think is this line, takes new location in the memory ???
Okay let's check by displaying hashcode of the s1 object which we created
System.out.println(s1.hashCode());
okay, let's check below code
String s2=new String("Kumar");
System.out.println(s2.hashCode()); // why this gives the different address ??
Okay, check this below code at last
String s3=new String("KUMAR");
System.out.println(s3.hashCode()); // again different address ???
YES, if you see Strings 's' and 's1' having the same hashcode because the value hold by 's' & 's1' are same that is 'kumar'
Let's consider String 's2' and 's3' these two Strings hashcode appears to be different in the sense, they both stored in a different location because you see their values are different.
since s and s1 hashcode is same because those values are same and storing in the same location.
Example 1:
Try below code and analyze line by line
public class StringImmutable {
public static void main(String[] args) {
String s="java";
System.out.println(s.hashCode());
String s1="javA";
System.out.println(s1.hashCode());
String s2=new String("Java");
System.out.println(s2.hashCode());
String s3=new String("JAVA");
System.out.println(s3.hashCode());
}
}
Example 2: Try below code and analyze line by line
public class StringImmutable {
public static void main(String[] args) {
String s="java";
s.concat(" programming"); // s can not be changed "immutablity"
System.out.println("value of s "+s);
System.out.println(" hashcode of s "+s.hashCode());
String s1="java";
String s2=s.concat(" programming"); // s1 can not be changed "immutablity" rather creates object s2
System.out.println("value of s1 "+s1);
System.out.println(" hashcode of s1 "+s1.hashCode());
System.out.println("value of s2 "+s2);
System.out.println(" hashcode of s2 "+s2.hashCode());
}
}
Okay, Let's look what is the difference between mutable and immutable.
mutable(it change) vs. immutable (it can't change)
public class StringMutableANDimmutable {
public static void main(String[] args) {
// it demonstrates immutable concept
String s="java";
s.concat(" programming"); // s can not be changed (immutablity)
System.out.println("value of s == "+s);
System.out.println(" hashcode of s == "+s.hashCode()+"\n\n");
// it demonstrates mutable concept
StringBuffer s1= new StringBuffer("java");
s1.append(" programming"); // s can be changed (mutablity)
System.out.println("value of s1 == "+s1);
System.out.println(" hashcode of s1 == "+s1.hashCode());
}
}
Any further questions?? please write on...
I modified the code of william with a output comments for better understandable
static void changeStr(String in) {
in = in+" changed";
System.out.println("fun:"+in); //value changed
}
static void changeStrBuf(StringBuffer in) {
in.append(" changed"); //value changed
}
public static void main(String[] args) {
StringBuffer sb = new StringBuffer("value");
String str = "value";
changeStrBuf(sb);
changeStr(str);
System.out.println("StringBuffer: "+sb); //value changed
System.out.println("String: "+str); // value
}
In above code , look at the value of str in both main() and changeStr() , even though u r changing the value of str in changeStr() it is affecting only to that function but in the main function the value is not changed , but it not in the case of StringBuffer..
In StringBuffer changed value is affected as a global..
hence String is immutable and StringBuffer is mutable...
In Simple , whatever u changed to String Object will affecting only to that function By going to String Pool. but not Changed...
A mutable variable is one whose value may change in place, whereas in an immutable variable change of value will not happen in place. Modifying an immutable variable will rebuild the same variable.
Mutable means you will save the same reference to variable and change its contents but immutable you can not change contents but you will declare new reference contains the new and the old value of the variable
Ex
Immutable -> String
String x = "value0ne";// adresse one
x += "valueTwo"; //an other adresse {adresse two}
adresse on the heap memory change.
Mutable -> StringBuffer - StringBuilder
StringBuilder sb = new StringBuilder();
sb.append("valueOne"); // adresse One
sb.append("valueTwo"); // adresse One
sb still in the same adresse i hope this comment helps
I Think I broke this down to the most basic form. If not then I apologize and will try to edit it. Why in my while loop do I need to cast String for the FirstName if I already set the variable to be a String? Am I doing something wrong? I am try to set the FirstName variable to be equal to the first token which is supposed to be the first word of the text file etc.
try
{
BufferedReader infileCust =
new BufferedReader(new FileReader("C:\\custDat.txt"));
}
catch(FileNotFoundException fnfe)
{
System.out.println(fnfe.toString());
}
catch(IOException ioe)
{
System.out.println(ioe.toString());
}
}
public static void createCustomerList(BufferedReader infileCust,
CustomerList custList) throws IOException
{
String FirstName;
String LastName;
int CustId;
//take first line of strings before breaking them up to first last and cust ID
String StringToBreak = infileCust.readLine();
//split up the string with string tokenizer
StringTokenizer st = new StringTokenizer(StringToBreak);
while(st.hasMoreElements()){
FirstName = (String) st.nextElement();
LastName = (String) st.nextElement();
CustId = Integer.parseInt((String) st.nextElement());
}
Edit: Apprently I can use nextToken() instead. But how come my variables are being shown as not used? Are they not within the scope of the while loop?
This is because nextElement returns Object. If you call nextToken, you would not need to cast.
From the documentation:
public Object nextElement()
Returns the same value as the nextToken method, except that its declared return value is Object rather than String. It exists so that this class can implement the Enumeration interface.
EDIT Regarding the variables that are not used: the reason you get the warning is that the variables are assigned, but not printed, saved, or analyzed in some way. If you add a call to, say, writeln with first and last name, the warnings would go away.
Because StringTokenizer.nextElement returns a java.lang.Object. See the docs here: http://docs.oracle.com/javase/1.4.2/docs/api/java/util/StringTokenizer.html You can use nextToken() : String instead if you prefer.
Because StringTokenizer.nextElement() returns
The same value as the nextToken method, except that its
declared return value is Object rather than String.It exists so that this class can implement the Enumeration interface.
How do String objects work in Java? How does term "immutable" exactly apply to string objects? Why don't we see modified string after passing through some method, though we operate on original string object value?
a String has a private final char[] . when a new String object is created, the array is also created and filled. it cannot be later accessed [from outside] or modified [actually it can be done with reflection, but we'll leave this aside].
it is "immutable" because once a string is created, its value cannot be changed, a "cow" string will always have the value "cow".
We don't see modified string because it is immutable, the same object will never be changed, no matter what you do with it [besides modifying it with reflection]. so "cow" + " horse" will create a new String object, and NOT modify the last object.
if you define:
void foo(String arg) {
arg= arg + " horse";
}
and you call:
String str = "cow";
foo(str);
the str where the call is is not modified [since it is the original reference to the same object!] when you changed arg, you simply changed it to reference another String object, and did NOT change the actual original object. so str, will be the same object, which was not changed, still containing "cow"
if you still want to change a String object, you can do it with reflection. However, it is unadvised and can have some serious side-affects:
String str = "cow";
try {
Field value = str.getClass().getDeclaredField("value");
Field count = str.getClass().getDeclaredField("count");
Field hash = str.getClass().getDeclaredField("hash");
Field offset = str.getClass().getDeclaredField("offset");
value.setAccessible(true);
count.setAccessible(true);
hash.setAccessible(true);
offset.setAccessible(true);
char[] newVal = { 'c','o','w',' ','h','o','r','s','e' };
value.set(str,newVal);
count.set(str,newVal.length);
hash.set(str,0);
offset.set(str,0);
} catch (NoSuchFieldException e) {
} catch (IllegalAccessException e) {}
System.out.println(str);
}
From the tutorial:
The String class is immutable, so that once it is created a String object cannot be changed. The String class has a number of methods, some of which will be discussed below, that appear to modify strings. Since strings are immutable, what these methods really do is create and return a new string that contains the result of the operation.
Strings in Java are immutable (state cannot be modified once created). This offers opportunities for optimization. One example is string interning, where string literals are maintained in a string pool and new String objects are only created if the particular string literal doesn't already exist in the pool. If the string literal already exists, a reference is returned. This can only be accomplished because strings are immutable, so you don't have to worry that some object holding a reference will change it.
Methods that appear to modify a string actually return a new instance. One example is string concatenation:
String s = "";
for( int i = 0; i < 5; i++ ){
s = s + "hi";
}
What actually happens internally (the compiler changes it):
String s = "";
for( int i = 0; i < 5; i++ ){
StringBuffer sb = new StringBuffer();
sb.append(s);
sb.append("hi");
s = sb.toString();
}
You can clearly see that new instances are created by the toString method (note that this can be made more efficient by directly using StringBuffers). StringBuffers are mutable, unlike Strings.
Every object has state. The state of a String object is the array of characters that make up the String, for example, the String "foo" contains the array ['f', 'o', 'o']. Because a String is immutable, this array can never be changed in any way, shape, or form.
Every method in every class that wants to change a String must instead return a new String that represents the altered state of the old String. That is, if you try to reverse "foo" you will get a new String object with internal state ['o', 'o', 'f'].
I think this link will help you to understand how Java String really works
Now consider the following code -
String s = "ABC";
s.toLowerCase();
The method toLowerCase() will not change the data "ABC" that s contains. Instead, a new String object is instantiated and given the data "abc" during its construction. A reference to this String object is returned by the toLowerCase() method. To make the String s contain the data "abc", a different approach is needed.
Again consider the following - s = s.toLowerCase();
Now the String s references a new String object that contains "abc". There is nothing in the syntax of the declaration of the class String that enforces it as immutable; rather, none of the String class's methods ever affect the data that a String object contains, thus making it immutable.
I don't really understood your third question. May be providing a chunk of code and telling your problem is a better option. Hope this helps.
You can also look into this blogpost for more understanding
[code samples are taken from the wiki. you can also look in there for more information]
So I want to be able to have a collection of mutable Strings in Java.
I have this test class to see the functionality of immutable Strings:
public class GetStringTest
{
private Vector<String> m_stringList;
public GetStringTest()
{
m_stringList = new Vector<String>();
m_stringList.add("zero");
m_stringList.add("one");
m_stringList.add("two");
m_stringList.add("three");
m_stringList.add("four");
m_stringList.add("five");
m_stringList.add("six");
}
public String getString(int index)
{
return m_stringList.get(index);
}
public String toString()
{
String str = "";
for (String item : m_stringList)
{
str += item + "\n";
}
return str;
}
public static void main(String[] args)
{
GetStringTest gst = new GetStringTest();
System.out.println("=== original content ===");
System.out.println(gst);
String strToChange = gst.getString(2); // "two"
strToChange = "eleventy-one";
System.out.println("=== with change ===");
System.out.println(gst);
}
}
The following is the output:
=== original content ===
zero
one
two
three
four
five
six
=== with change ===
zero
one
two
three
four
five
six
What can I do to store these Strings as mutable? I was thinking of having a StringObject class that would simply contain a reference to a String. Is this the best option?
StringBuilder is a mutable string representation.
A mutable sequence of characters.
It is the unsynchronized (and more efficient) version of StringBuffer
May I recommend you to use a StringBuffer.
From the API:
A thread-safe, mutable sequence of characters. A string buffer is like a String, but can be modified. At any point in time it contains some particular sequence of characters, but the length and content of the sequence can be changed through certain method calls.
Note that if you don't plan to use it in you're better off with a StringBuilder as it is unsynchronized and faster.
If you want to change (replace) a list element you could add
public String setString(int index,String str)
{
return m_stringList.set(index, str);
}
And call:
String strToChange = gst.setString(2,"eleventy-one");
instead of:
String strToChange = gst.getString(2); // "two"
strToChange = "eleventy-one";
Thats because your Vector holds a reference to an immutable String, the reference can be replaced by a another String, the String itself can't be changed.
Ther's not a reference copy in Java. You can do a new method to GetString class to set a new String to a determined index.
Example:
gst.changeString(index, "NewString");
changeString internally set reference to "NewString".