String vs new String

[bugger]

Code:

String myString = “hello world”;
String myString1 = new String(“Hello World”);

Whats the difference between the statements above?

[m_srikanth]

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. For example:


String str = "abc";
is equivalent to:


char data[] = {'a', 'b', 'c'};
String str = new String(data);

[Eranga]

Yep, it is so useful when you want to change the string dynamically.

[bugger]

Thanks for the explanation. So it means that when we create a String using the following syntax, we create String buffer that can be changed unlike String constants.

Code:

String str = new String("String Msg");

[Eranga]

Correct. Size of the buffer same as the number of characters int the string.

[javaplus]

Ok. If the buffer has the same size, as the number of characters in the string, how about the followinf example:

Code:

String str = new String("Austarlia");
str = "Australia Football Team";

str was declared with buffer size 9. Then I put a String of size 23 characters. Of course its not a syntax error but does this mean that the buffer is flexible and grow according to the String??

[Eranga]

Try it and see.

Simply print some characters or find the size of the buffer. Then you can see what's going on there.

[javaplus]

I tried the following:

Code:

String myString = new String("Hello");
System.out.println("Buffer size: " + myString.length());
myString = "Hello-Hello-Hello";
System.out.println("Buffer size: " + myString.length());
myString = "hi";
System.out.println("Buffer size: " + myString.length());

Output:

Code:

Buffer size: 5
Buffer size: 17
Buffer size: 2

So buffer grows or shrink according to the String. Makes sense

[Eranga]

Yep, now you know what is happened there. That is the important of String objects.

[bugger]

Thanks all of you. Now I know the difference.

[Eranga]

Ok. It's cool, isn't it?

[javaplus]

Well, I discovered something interesting. Review the code below. I think, every time we change the value in String, new object is being created.

Code:

String myString = new String("Hello");
System.out.println("Buffer size: " + myString.length() + 
" Hash code " + myString.hashCode());

myString = "Hello-Hello-Hello";
System.out.println("Buffer size: " + myString.length() + 
" Hash code " + myString.hashCode());

myString = "hi";
System.out.println("Buffer size: " + myString.length() + 
" Hash code " + myString.hashCode());

Output:

Code:

Buffer size: 5 Hash code 69609650
Buffer size: 17 Hash code -2022942916
Buffer size: 2 Hash code 3329

[Eranga]

I don't think so. It use the same object with different hash code, that mean different memory locations. I'm not 100% sure, I want to check it now. May be some explanation can be available in documentation.

[javaplus]

Yes, may be it stores at different memory location as the size changes.
Please share if you have some interesting information.

[Eranga]

Sure. As far as I know the effect is that dynamically changing the buffer, first what have happened is destroyed/delete the first buffer, then create a new buffer with new size. So the memory location is differ. Basically memory location is selected randomly.

If you can just test your code on another machine. I think it give the different hash code for you.

[javaplus]

Makes sense. But if the buffer is deleted and then new is created, isnt it like

Code:

String a = "a";
a= "aa";
...

I mean then what is the real difference?

[Eranga]

Real difference comes with real application programming. Working with objects gives the best way to work. That is why OOP is handy.

[javaplus]

Yep - OOP sure is handy.

[Eranga]

I have no a best explanation to you on this. Actually you can do above all the things we have tested by using String constant. Depends on the situation of the coding it should be decide, either working with constants or with objects.

[bugger]

Thanks guys for the fruitful discussion.
I learned new things.

Keep educating.