a highly recommended mock site: http://www.jiris.com/mock

[Mellihoney Michael]

Here is a question in it:
class Test001
{
int i;
public Test001(int i) { this.i = i; }
public String toString()
{
if(i == 0) return null;
else return "" + i;
}
public static void main(String[ ] args)
{
Test001 t1 = new Test001(0);
Test001 t2 = new Test001(2);
System.out.println(t1);
System.out.println(t2);
}
}
why does it compiled, but thrown an exception in Runtime.

[Pradeep bhatt]

It is as good as writing
System.out.print(null)
which gives compile error

[Sarma Lolla]

Any method that returns a reference to Object can always return null.
How ever when the toString method is called in System.out.println() the method is getting called on null referrence and hence the run time error.

[Mellihoney Michael]

Pradeep Bhat,Sarma Lolla :
thank you for clearing.
by the way what do you think of this mock exam?

[Mellihoney Michael]

class Test001
{
int i;
public Test001(int i) { this.i = i; }
public String toString()
{
if(i == 0) return null;
else return "" + i;
}
public static void main(String[ ] args)
{
Test001 t1 = new Test001(0);
Test001 t2 = new Test001(2);
System.out.println(t1.toString());
System.out.println(t2);
}
}
is this compiled without an error, and print "null" to standard output
[ February 18, 2003: Message edited by: Mellihoney Michael ]

[Erico Doon]

class Test001
{
int i;
public Test001(int i) { this.i = i; }
public String toString()
{
if(i == 0) return null;
else return "" + i;
}
public static void main(String[ ] args)
{
Test001 t1 = new Test001(0);
Test001 t2 = new Test001(2);
System.out.println(t1);
System.out.println(t2.toString( ));
}
}
compile successfully
runtime exception

[Mellihoney Michael]

ERIC:
brother,it indicated above two code fragments are identical!

[ February 18, 2003: Message edited by: Mellihoney Michael ]

[Garrett Smith]

I'm not content with my understanding of what I see.
Why does System.out.println(t1) have a NullPointerException?

print
public void print(Object obj)
Print an object. The string produced by the String.valueOf(Object) 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.
Parameters:
obj - The Object to be printedS
See Also:
Object.toString()

Now if I use:

    public static void main(String[ ] args) {
    
        
        PrintNull2 t1 = new PrintNull2(0);
        PrintNull2 t2 = new PrintNull2(2);
        
        System.out.println(String.valueOf(t1));
        System.out.println(t2);
    }

I have no problem. It somehow seems that print (or println) isn't doing what the Javadoc says it's supposed to do.

[Prashant Neginahal]

Here ,I did'nt understand
Why does System.out.println(t1) have a NullPointerException?

[Mellihoney Michael]

it is really a controversy one!
but the fact is that
it is compiled OK,and throws the Exception on RunTime!

[John Lee]

Thanks, this is a good site!

[Tom Adams]

System.out.println(t1); // automagically becomes
System.out.println( String.valueOf(t1) );
The String.valueOf says...
if the argument is null, then a string equal to "null"; otherwise, the value of obj.toString() is returned
In our case t1 is not null so obj.toString() is called which is cleverly overridden.
The t1.toString() returns a null value...then System.out.println executes thinking it has a reference to a String. It points to nothing so...NullPointerException (Thrown when an application attempts to use null in a case where an object is required.)

  
public class NullException {
public static void main(String[] args) {
System.out.println( new Object() { public String toString() { return null; } });
}
}

[ February 19, 2003: Message edited by: Tom Adams ]

[Garrett Smith]

Thanks.
I guess the moral to the story is "When overriding Object.toString(), don't return null; return a String."
Taking a closer look at the stack trace:

Exception in thread "main" java.lang.NullPointerException
        at java.io.Writer.write(Writer.java:126)
        at java.io.PrintStream.write(PrintStream.java:303)
        at java.io.PrintStream.print(PrintStream.java:462)
        at java.io.PrintStream.println(PrintStream.java:599)
        at PrintNull2.main(PrintNull2.java:18)

So the error occured at Writer.write. I am guessing the signature is the one below, but I don't know. I don't have Writer.java (source file).

write
public void write(char[] cbuf)
throws IOException
Write an array of characters.
Parameters:
cbuf - Array of characters to be written
Throws:
IOException - If an I/O error occurs

Maybe it's trying to loop through a null array:

for(int i = 0; i < cbuf.length; i++) // cbuf is null.

[Karin Paola Illuminate]

A difficult question. Thanks. This will help with studying for the exam.
But, unfortunately, I still don't understand the answer to the question. Does this mean that a null String reference can be passed to the println() method, but a null literal may not be passed to the method?
Do I have to know the exact reason for the exam? Because, I don't know the exact reason for this.
Thanks.
My Study Notes
[ March 19, 2003: Message edited by: Karin Paola Illuminate ]

[Lih Chang]

The following codes copied form the java source file may give you a clue what's happening:
1: public void println(Object x) {
synchronized (this) {
print(x);
newLine();
}
2: the write method which in turn calls the write method in Writer class
public void print(Object obj) {
write(String.valueOf(obj));
}
3: the obj passed in is Test001(0) which is of course not null, if the obj passed in was null, it will then printed out as null.
public static String valueOf(Object obj) {
return (obj == null) ? "null" : obj.toString();
}
4: when calling obj.toString() as in step 3, it returns null as in this case. By calling str.length() in the write method, it will of course throw NullPointerException, since it is same as calling null.length() and length() is an instance method not a class method.
//write mehtod from Writer class
public void write(String str) throws IOException {
write(str, 0, str.length());
}

Hope this help, please forgive my poor English.