Will a new String("something") reference placed in literal pool?

[Nithya Vasudevan]

Hello

I went through this documentation on Strings http://www.javaranch.com/journal/200409/Journal200409.jsp and I would like to know what happens when there is a statement like this,

String s = new String("something");

Now, will the string literal pool contain a reference to string "something" object in heap ?

Thanks

[Winston Gutkowski]

Nithya Vasudevan wrote:Now, will the string literal pool contain a reference to string "something" object in heap ?

Yes.

Winston

[Campbell Ritchie]

To elaborate on what Winston wrote. Yes. And no.
To the question in your post, yes. To the question in the thread title, no. Which just shows how careful you have to be about what you ask. Try adding this line

System.out.printf("The String \"something\" is in the String pool: %b", s == "something");

[Jeff Verdegan]

Nithya Vasudevan wrote:Hello

I went through this documentation on Strings http://www.javaranch.com/journal/200409/Journal200409.jsp and I would like to know what happens when there is a statement like this,

String s = new String("something");

Now, will the string literal pool contain a reference to string "something" object in heap ?

Before you even get to that line of code, the pool has the "something" literal in it. It's put there when the class is loaded, if there wasn't already a "something" String there.

Then, when you get to that line, a new String object is created, and the constructor invoked is the one that takes a reference to an existing String that will be copied. We pass it a reference to the "something" String in the pool, that String object is copied, and we now have two "something" String objects, one in the pool, and one not.

[Dennis Deems]

In your example, a String literal will be placed in the pool, but only because the literal is passed into the constructor. Be aware that we can use new String without a literal.

Consider this:

char[] chars = {'a','n','y','t','h','i','n','g'};
String g = new String(chars);
byte[] bytes = {'a','n','y','t','h','i','n','g'};
String y = new String(bytes);
System.out.println( y == g);
System.out.println(g + " equals " + y + "? " + g.equals( y ));

Here, neither String is placed in the literal pool.

[Nithya Vasudevan]

Thanks all for clarifying.