Question Regarding Strings

[Gary Farms]

I understand that strings are immutable. Once a string is given a value, it's value can never change. However, when the following small program is run, it prints true. Why? Doesn't s2.toUpperCase() change s2's original value? Here's the program:
class A
{
public static void main(String[] args)
{
String s1="HELLO WORLD";
String s2="hello world";
if(s1.equals(s2.toUpperCase()) == true)
System.out.println("true");
}
}

[James O'Dell]

Gary,
In your code "s2.toUpperCase()" returns a brand new String object. It doesn't change the original s2.
--Jim

[Sumanta Mandal]

s2.toUpperCase() does not change s2's original value. You may test this by printing System.out.print(s2); it after you invoke s2.toUpperCase(). In this case you are testing the equivalence with the returned value by the method toUpperCase().

[Anthony Villanueva]

Originally posted by Gary Farms:
Doesn't s2.toUpperCase() change s2's original value?

No. This method returns a new String that has the changes. Try this:

public class Test44
{   
public static void main(String[] arg) 
{
String s1="HELLO WORLD";
String s2="hello world";
if(s1.equals(s2.toUpperCase()) == true) System.out.println("true");
System.out.println(s2);
}
}

[Dave Landers]

No. s2.toUpperCase() returns a brand new string that is an upper case'ed copy of s2. s2 is unchanged.
Try this:

 
class A
{
    public static void main(String[] args)
    {
        String s1="HELLO WORLD";
        String s2="hello world";
        String s3 = s2.toUpperCase();
        if(s1.equals(s3) == true)
            System.out.println("true");
        System.out.println( s2 );
        System.out.println( s3 );
    }
} 

Even if you do
s2 = s2.toUpperCase();
You still get a new object. The toUpperCase method creates a new String object and you assign that new reference to the thing called s2. And s2 now points to a different object. The original object (containing "hello world") is unchanged.

[Dirk Schreckmann]

Wow! Four correct answers at the same time. Look out!