What I read about strings is when a string object is created in Java it is immutable. For example:
String s=new String();
s="abc";
s="xyz";
Does the String s no longer point to "abc"?
And another thing: what is size of s;
Is the String object analogous to a char* pointer in C in terms of the memory model?
No, Java String is not as char* in C. If you are looking for the analogue char[] in java is something like this.
String is a class that wraps char array and provides a lot of functionality. Moreover it is immutable, i.e. you cannot change its content. You can only create another string. Additionally String is final, so you cannot subclass it.
String is special class. Only string supports operator (+). All other classes do not support operators at all, even primitive wrappers (Integer, Double etc). Presence of string constant in code "foobar" invokes java.lang.String constructor.
Yes, Strings in Java are immutable. That is to say, the content pointed too by a string variable cannot be changed after it has been initialized. Using your own examples to illustrate:
String s=new String();
In the above code, you have created a new, empty String and assigned it to the s variable.
s="abc";
You have created another new String, this time with content "abc". You have set the variable s to point to this new String. Your previous, empty string will get garbage collected at some point in the future.
s="xyz";
Similar to above. You have created another new String, with content "xyz", and set the variable s to point to it. The previous "abc" string will get garbage collected at some point in the future.
Note that at no point did you actually modify the empty string to become "abc", or modify the "abc" string to become "xyz". All operations on String that concatenate, convert case, or otherwise appear to modify the String actually return a new String with the function results. To illustrate:
String s = new String("Hello");
String b = s.concat(" World");
System.out.println(s); // This will NOT produce 'Hello World'
System.out.println(b); // Whereas, this will
As to the last part of your question, you would use the length() function to determine the 'size' of your string. Note that this is based on Unicode encoding, and might not be what you expect the length to be as defined in ANSI C.
Que: Does the String s no longer point to "abc"?
Ans: No, it point to "xyz" regarding your code.
Que: what is size of s?
Ans: Its a bit tricky, "s" is the reference to "xyz". It is true for any declared variables which are referencing an object. So it is better to say, s is pointing to "xyz".
One important hint, memory of an object & referencing an object takes different memory space.
Immutable: A object is immutable if you cant change its property after instantiation.
Related
Javadoc said that:
The String class represents character strings. All string literals in Java programs, such as "abc", are implemented as instances of this class.
We know that String class has two properties namely: value[] and hash, and String literal are stored in a String pool.
But I am not able to figure out how that String literal is initialized before putting to that pool. As if I debug the string literal later, I can see the value[] and hash is somehow populated.
Does JVM invoke a special instruction?
JVM creates a new string literal object during constant pool resolution, if the same string has not been put to the string table before by String.intern call.
It is not specified how JVM creates and initializes such strings, so JVM may do whatever it wants as long as the result object is a regular java.lang.String instance that can be accessed from application code.
As to HotSpot JVM, it does not call any of String constructors, it just allocates a new object in Heap and fills in the fields in C++ code, see java_lang_String::basic_create:
Handle java_lang_String::basic_create(int length, TRAPS) {
assert(initialized, "Must be initialized");
// Create the String object first, so there's a chance that the String
// and the char array it points to end up in the same cache line.
oop obj;
obj = InstanceKlass::cast(SystemDictionary::String_klass())->allocate_instance(CHECK_NH);
// Create the char array. The String object must be handlized here
// because GC can happen as a result of the allocation attempt.
Handle h_obj(THREAD, obj);
typeArrayOop buffer;
buffer = oopFactory::new_charArray(length, CHECK_NH);
// Point the String at the char array
obj = h_obj();
set_value(obj, buffer);
// No need to zero the offset, allocation zero'ed the entire String object
assert(offset(obj) == 0, "initial String offset should be zero");
//set_offset(obj, 0);
set_count(obj, length);
return h_obj;
}
hash field of such new object is initialized to zero. The right hash code will be calculated on the first call to String.hashCode.
Does JVM invoke a special instruction?
No. There is no "special" JVM instruction to do that.
The classloading infrastructure will most likely be creating the String objects that correspond to the literals using one of the String constructors; e.g. String(char[]) or String(byte[]). It will be getting the characters or bytes from the "constant pool" area in the ".class" file.
Follow breakpoint on public String(char value[]) of String class.
Edit: Whoever downvoted this:
Firstly, did you try as per answer? I had tried and it stops exactly in that method. I had tried with two different IDEs.
Secondly, value member(as pointed in OP) is assigned in that method. If you indeed tried, you would have got that. Thirdly, hash member(as pointed in OP) is assigned a value on the call to hashCode() and hash member is assigned a value in it.
Clearly, you haven't tried. If you did, investigation on breakpoint would have led you to answers in OP:
1) how that String literal is initialized
and
2) ...somehow populated
.
In Java, apparently, String s = "foo" is preferred over String s = new String("foo").
Why? Isn't a new string object created in both cases? Why would the first case preclude calling a constructor?
Why?
Because the second approach results in two string objects (the original due to the string literal, plus an explicit copy).
The first case is a string literal, simply a shorthand the language offers you to create a string. The String class constructor still gets called, just not explicitly, which means less typing and less code clutter.
The second case takes the String object already created by the literal and passes it to a constructor, which copies the content to create a new, separate String object. The literal will still be around because literals are interned.
There is rarely a point to using the String constructor (pretty much only when you've created a substring of a very large string and want to release the memory used by the rest of the string, because substrings by default use the same underlying char array as the original string, just with a different offset and length.
I don't think it's preferable. I assume the only "benefit" you get is that if you wrongfully use the "==" operator rather than the equals method, have two different instances of a string will fail faster which will prompt you to fix your code. (the == operator may "succeed" and fail unpredictably)
Unless of course your code requires you to construct two different instances for whatever reason
Why? Isn't a new string object created in both cases?
No, the initial form being a string literal will be interned such that only one instance is created:
String s = "foo";
String s2 = "foo";
s == s2 => true
If Strings are immutable in Java, then how can we write as:
String s = new String();
s = s + "abc";
Strings are immutable.
That means that an instance of String cannot change.
You're changing the s variable to refer to a different (but still immutable) String instance.
Your string variable is NOT the string. It's a REFERENCE to an instance of String.
See for yourself:
String str = "Test String";
System.out.println( System.identityHashCode(str) ); // INSTANCE ID of the string
str = str + "Another value";
System.out.println( System.identityHashCode(str) ); // Whoa, it's a different string!
The instances the str variable points to are individually immutable, BUT the variable can be pointed to any instance of String you want.
If you don't want it to be possible to reassign str to point to a different string instance, declare it final:
final String str = "Test String";
System.out.println( System.identityHashCode(str) ); // INSTANCE ID of the string
str = str + "Another value"; // BREAKS HORRIBLY
The first answer is absolutely correct. You should mark it as answered.
s = s+"abc" does not append to the s object. it creates a new string that contains the characters from the s object (of which there are none) and "abc".
if string were mutable. it would have methods like append() and other such mutating methods that are on StringBuilder and StringBuffer.
Effective Java by Josh Bloch has excellent discussion on immutable objects and their value.
Immutable Classes are those whose methods can change their fields, for example:
Foo f = new Foo("a");
f.setField("b"); // Now, you are changing the field of class Foo
but in immutable classes, e.g. String, you cannot change the object once you create it, but of course, you can reassign the reference to another object. For example:
String s = "Hello";
s.substring(0,1); // s is still "Hello"
s = s.substring(0,1); // s is now referring to another object whose value is "H"
String s = new String();
Creates a new, immutable, empty string, variable "s" references it.
s = s+"abc";
Creates a new, immutable, string; the concatenation of the empty string and "abc", variable "s" now references this new object.
Just to clarify, when you say s = s+"abc";
That means, create a new String instance (which is composed of s and "abc") then assign that new String instance to s. So the new reference in s is different from the old.
Remember, a variable is effectively a reference to an object at some specific memory location. The object at that location stays at that location, even if you change the variable to refer to a new object at a different location.
String s = new String();
An empty String object ("") is created. And the variable s refers to that object.
s = s + "abc";
"abc" is a string literal (which is nothing but a String object, which is implicitly created and kept in a pool of strings) so that it can be reused (since strings are immutable and thus are constant). But when you do new String() is totally different because you are explicitly creating the object so does not end up in the pool. You can throw is in the pool by something called interning.
So, s + "abc" since at this point concatenation of and empty string ("") and "abc" does not really create a new String object because the end result is "abc" which is already in the pool. So, finally the variable s will refer to the literal "abc" in the pool.
I believe you are all making this much more complicated than it needs to be, and that simply confuses people who are trying to learn!
The primary benefit of making an object immutable in Java is that it can be passed by reference (e.g. to another method or assigned using the assignment operator) without having to worry about downstream changes to the object causing issues in the current method or context. (This is very different than any conversation about the thread safety of an object.)
To illustrate, create an application that passes a String as a parameter to a separate method, and modify the String in that method. Print the String at the end of the called method and then after control returns to the calling method. The Strings will have different values, and that's because they point to different memory locations, a direct result of "changing" the immutable String (creating a new pointer and pointing it to a new value behind the scenes). Then create an application that does the same things except with StringBuffer, which is not immutable. (For example, you can append to the StringBuffer to modify it.) The printed StringBuffers will have the same values, and that is because it is (a) being passed by reference, as Java does with all objects passed to methods as parameters and (b) mutable.
I hope this helps folks who are reading this thread and trying to learn!
this is probably an easy question but not for me at this stage.
I read that (I quote) "All instances of the String class are constants, meaning they cannot be changed after they have been created. But strings can be replaced".
This is confusing. What does it mean that a String variable cannot be changed but the string can?
Can anyone please give me code example where we would try to change the variable but fail?
Cheers for the help.
When you write:
String myString = "Hi!";
You're doing two things. The first one is defining a reference called myString the second one is creating a String Object. That String Object contains "Hi!", and there isn't a way to change that. In other words, there isn't a set method to change the string:
myString.set("another content");
However, you can create a new Object and change your myString reference to point to it. The important thing to get is that your myString isn't the String Object itself but just a reference to it.
myString = "New content";
When you do that, the old String is not pointed by any variable any more and is a candidate for garbage collection. Also any other operation on the String, such as substring, uppercase, etc. will create a new String Object.
When an Object can't be changed after being created is called Immutable. In Java Strings are not only immutable, but also final, so that you can't subclass a String to change its behaviour.
Strings aren't constants, they're immutable, which is something else altogether. It means that you cannot modify the contents of string object, but you can make a new one based on the first, e.g.
String hw = "hello world";
hw.ToUpperCase();
The latter doesn't change the original string hw but creates a new string with all characters to uppercase.
What does it mean that a String variable cannot be changed but the string can?
Actually the text that you quoted means the exact opposite of that.
A String variable can be assigned to (i.e. replaced). Example is:
String s = "first";
s = "second";
This replaces the reference to "first" with a reference to "second".
I can't give you a legitimate example of changing the contents / state of a String because you can't do it*.
* Actually, you can do it by breaking a String object's encapsulation using reflection. But it is a really, really bad idea.
The answer to the question is this: in Java string literal is an immutable object.
As previously stated by fellow programmers - you may change the reference, yes, but you may never change an immutable object (it is guaranteed not to be possible to change it).
While looking at online code samples, I have sometimes come across an assignment of a String constant to a String object via the use of the new operator.
For example:
String s;
...
s = new String("Hello World");
This, of course, compared to
s = "Hello World";
I'm not familiar with this syntax and have no idea what the purpose or effect would be.
Since String constants typically get stored in the constant pool and then in whatever representation the JVM has for dealing with String constants, would anything even be allocated on the heap?
The one place where you may think you want new String(String) is to force a distinct copy of the internal character array, as in
small=new String(huge.substring(10,20))
However, this behavior is unfortunately undocumented and implementation dependent.
I have been burned by this when reading large files (some up to 20 MiB) into a String and carving it into lines after the fact. I ended up with all the strings for the lines referencing the char[] consisting of entire file. Unfortunately, that unintentionally kept a reference to the entire array for the few lines I held on to for a longer time than processing the file - I was forced to use new String() to work around it, since processing 20,000 files very quickly consumed huge amounts of RAM.
The only implementation agnostic way to do this is:
small=new String(huge.substring(10,20).toCharArray());
This unfortunately must copy the array twice, once for toCharArray() and once in the String constructor.
There needs to be a documented way to get a new String by copying the chars of an existing one; or the documentation of String(String) needs to be improved to make it more explicit (there is an implication there, but it's rather vague and open to interpretation).
Pitfall of Assuming what the Doc Doesn't State
In response to the comments, which keep coming in, observe what the Apache Harmony implementation of new String() was:
public String(String string) {
value = string.value;
offset = string.offset;
count = string.count;
}
That's right, no copy of the underlying array there. And yet, it still conforms to the (Java 7) String documentation, in that it:
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.
The salient piece being "copy of the argument string"; it does not say "copy of the argument string and the underlying character array supporting the string".
Be careful that you program to the documentation and not one implementation.
The only time I have found this useful is in declaring lock variables:
private final String lock = new String("Database lock");
....
synchronized(lock)
{
// do something
}
In this case, debugging tools like Eclipse will show the string when listing what locks a thread currently holds or is waiting for. You have to use "new String", i.e. allocate a new String object, because otherwise a shared string literal could possibly be locked in some other unrelated code.
String s1="foo"; literal will go in StringPool and s1 will refer.
String s2="foo"; this time it will check "foo" literal is already available in StringPool or not as now it exist so s2 will refer the same literal.
String s3=new String("foo"); "foo" literal will be created in StringPool first then through string arg constructor String Object will be created i.e "foo" in the heap due to object creation through new operator then s3 will refer it.
String s4=new String("foo"); same as s3
so System.out.println(s1==s2); //true due to literal comparison.
and System.out.println(s3==s4);// false due to object comparison(s3 and s4 is created at different places in heap)
The sole utility for this constructor described by Software Monkey and Ruggs seems to have disappeared from JDK7.
There is no longer an offset field in class String, and substring always use
Arrays.copyOfRange(char[] original, int from, int to)
to trim the char array for the copy.
Well, that depends on what the "..." is in the example. If it's a StringBuffer, for example, or a byte array, or something, you'll get a String constructed from the data you're passing.
But if it's just another String, as in new String("Hello World!"), then it should be replaced by simply "Hello World!", in all cases. Strings are immutable, so cloning one serves no purpose -- it's just more verbose and less efficient to create a new String object just to serve as a duplicate of an existing String (whether it be a literal or another String variable you already have).
In fact, Effective Java (which I highly recommend) uses exactly this as one of its examples of "Avoid creating unnecessary objects":
As an extreme example of what not to do, consider this statement:
String s = new String("stringette"); **//DON'T DO THIS!**
(Effective Java, Second Edition)
Here is a quote from the book Effective Java Third Edition (Item 17: Minimize Mutability):
A consequence of the fact that immutable objects can be shared freely
is that you never have to make defensive copies of them (Item
50). In fact, you never have to make any copies at all because the
copies would be forever equivalent to the originals. Therefore, you
need not and should not provide a clone method or copy constructor
(Item 13) on an immutable class. This was not well understood in the
early days of the Java platform, so the String class does have a copy
constructor, but it should rarely, if ever, be used.
So It was a wrong decision by Java, since String class is immutable they should not have provided copy constructor for this class, in cases you want to do costly operation on immutable classes, you can use public mutable companion classes which are StringBuilder and StringBuffer in case of String.
Generally, this indicates someone who isn't comfortable with the new-fashioned C++ style of declaring when initialized.
Back in the C days, it wasn't considered good form to define auto variables in an inner scope; C++ eliminated the parser restriction, and Java extended that.
So you see code that has
int q;
for(q=0;q<MAX;q++){
String s;
int ix;
// other stuff
s = new String("Hello, there!");
// do something with s
}
In the extreme case, all the declarations may be at the top of a function, and not in enclosed scopes like the for loop here.
IN general, though, the effect of this is to cause a String ctor to be called once, and the resulting String thrown away. (The desire to avoid this is just what led Stroustrup to allow declarations anywhere in the code.) So you are correct that it's unnecessary and bad style at best, and possibly actually bad.
There are two ways in which Strings can be created in Java. Following are the examples for both the ways:
1) Declare a variable of type String(a class in Java) and assign it to a value which should be put between double quotes. This will create a string in the string pool area of memory.
eg: String str = "JAVA";
2)Use the constructor of String class and pass a string(within double quotes) as an argument.
eg: String s = new String("JAVA");
This will create a new string JAVA in the main memory and also in the string pool if this string is not already present in string pool.
I guess it will depend on the code samples you're seeing.
Most of the times using the class constructor "new String()" in code sample are only to show a very well know java class instead of creating a new one.
You should avoid using it most of the times. Not only because string literals are interned but mainly because string are inmutable. It doesn't make sense have two copies that represent the same object.
While the article mensioned by Ruggs is "interesting" it should not be used unless very specific circumstances, because it could create more damage than good. You'll be coding to an implementation rather than an specification and the same code could not run the same for instance in JRockit, IBM VM, or other.