Wats the Output...

[Shiaber Shaam]

public class ObjectTest {
String s;
public static void main(String args[]) {
ObjectTest o = new ObjectTest();
System.out.println(s);

}
}

[joshua antony]

http://www.jchq.net/certkey/0401certkey.htm--read this, most of your concepts will be cleared.
If you put effort yourself you will understand better than what we reply.

Regarding the output, you will get compilation error, since you are trying to use s before initializing it.(Local variables need to be initialised before use)

[Gunjaan Agarwal]

gives u compilation error as non-static field is being accessed in static method

[joshua antony]

oh sorry the output is null.

[Gunjaan Agarwal]

its not a local variable..its an instance variable..compilation error is coz no class instance is used to access s

[Bonny Sen]

But the code will run if you use System.out.println("Output: " + o.s);
else it will give a compiler error as a non-static variable cannot be referenced from a static context

[Shiaber Shaam]

I may be wrong in asking that qn.. let me review...

public class ObjectTest {
public static void main(String args[]) {
String s=null;
System.out.println(s);
}
}

Now i am clear with my qn.. i want to clarify whether it gives nullptrexcpn or just prints null... and with reasons....

[Amit Goyal]

when u prints an String object through System.out.println() method, toString() method is called.
Since there is no object i.e. String = null ; you will get NullPointerException.


Regards,

Amit Goyal

[Gunjaan Agarwal]

it will print null amit.Does not give a exception.An exception occurs when u don initaialize it.However if it was a instance var n not local u need not initialize it as instance var are automatically initialized.

[S Thiyanesh]

null will be printed

[Shiaber Shaam]

i want clarifications that it will give nullptrexcpn or prints null.. and also WHY???

[Shiaber Shaam]

i want clarifications that it will give nullptrexcpn or prints null.. and also WHY???

[Amit Goyal]

oopss! Gunjaan ur right.
Once i gone through the internal implementation of println() method i got the ans.

Thanx for correcting me!

Following the impl:

   
 
    /**
     * Returns the string representation of the <code>Object</code> argument.
     *
     * @param   obj   an <code>Object</code>.
     * @return  if the argument is <code>null</code>, then a string equal to
     *          <code>"null"</code>; otherwise, the value of
     *          <code>obj.toString()</code> is returned.
     * @see     java.lang.Object#toString()
     */
    public static String valueOf(Object obj) {
return (obj == null) ? "null" : obj.toString();
    }
 
 

Regards,

Amit Goyal

[Bonny Sen]

The variable String s is an instance of the class ObjectTest and it has not been explicitly initialized. Sowhen the class loads then s would be initialized to its corresponding default value, whic is null for strings. So null is printed when u call it as: System.out.println("Output: " + o.s);

[wise owen]

The PrintStream's "public println(Object x)" calls its "public void print(Object obj)" to print an object. The string produced by the "String.valueOf(Object obj)" method is translated into bytes according to the platform's default character encoding, and these bytes are written in exactly the manner of the write(int) method.


Here is the implementation of String.valueOf(Object obj):

public static String valueOf(Object obj)
{
  return obj == null ? "null" : obj.toString();
}

In your case, null string will be printed.

[Shiaber Shaam]

Here... Everything gets cleared... Thank U all.....

[Mark Spritzler]

"Bonny" and "Shaiber"

Change your display names to meet the JavaRanch Naming Policy of using your real first and real last names. Failing to comply can result if closing accounts.

Thanks for understanding.

Mark