Why strange output of this String question

hi folks !
please help me out to understand this question

Code First :
class StringDemo
{public static void main(String ss[])
{String s= "HELLO";
s.toLowerCase();
System.out.println(s);
}
}
its output is : HELLO

Code Second :
class StringDemo
{public static void main(String ss[])
{String s= "HELLO";
System.out.println(s.toLowerCase();

}
}
its output is : hello

why o/p is different in both cases, as we are not changing the reference to newly created string object

String Object is immutable. The "s.toLowerCase()" will return a new String object but the string referenced by s will not change.
[ May 11, 2006: Message edited by: wise owen ]

Read this thread for more detail.

hi,

System.out.println will print the returning value of the method.
in this case s.toLowerCase() is returning a new String and it is being printed in the console....

plz correct if i am wrong

regards
krishna bulusu

hi wise !

where the returned object by the method s.toLowerCase() in code A: goes

as in both cases references is not being cjanged

The created String Object will be lost because not variable to reference it.
This is portion of code in my previous link. It show the process step by step.

Example:

String s = "XYZ"; //1. a object created

s.concat("ABC"); //now a fresh copy of XYZ created and "ABC" to it and the
//reference to XYZABC returned, but it get lost
//as you are not assiging reference to it.
System.out.println(s); // its "XYZ"

now lets see if you write:

System.out.println(s.concat("ABC")); //it will print XYZABC but the s=XYZ

if you write:
s = s.concat("ABC"); //then s=XYZABC

Yup,what u need to do is this:
s=s.toLowerCase();