Thread: Please solve my problem

Originally Posted by Supun Pramoda

L1: It's default value of char(null character ('\0'))

Prints 3 null (0 value) characters. The println choosen by compiler is the println(char[] x) that correctly prints each character in the array, then a newline. Default value in char array elements is 0.

Note that in a real console (like that in Windows), you don't see a null character, it's invisible.

Originally Posted by Supun Pramoda

L2: first two array element when we create it in heap.

Prints characters 'A' then 'B', then a null character.

Originally Posted by Supun Pramoda

L3: I can't explain about that..please help me.

l+" "+cr is a string concatenation, which under the hood uses StringBuffer (or StringBuilder, from Java 5). On l and cr, the compiler invokes the toString() method. l and cr are arrays, and they derive from Object. However array does not redefine the toString(), so remains the toString() from Object, that can return only something like "[I@41675ec4" that is a "reference" of the array.

Originally Posted by Supun Pramoda

L4 : This is the same in L2.

Yes

Originally Posted by Supun Pramoda

L5: second element of array object..

That, at that moment, is 'B'

Originally Posted by Supun Pramoda

L6 : I can't explain about that one also...

0x12996d6 is an hexadecimal literal value. It's value, in decimal, is 19502806 (what is printed).

I'm reposting the code with your explanations. I'm glad we did that, because I know I didn't understand your question until you did:

class Example{
	public static void main(String args[]){
		char cr[]=new char[3];
		int l[]=new int[3];
		// L1: It's default value of char(null character ('\0'))
		System.out.println(cr);	//L1
		cr[0]='A';
		cr[1]='B';
		// L2: first two array element when we create it in heap.
		System.out.println(cr);		//L2
		l[0]=78;
		// L3: I can't explain about that..please help me.
		System.out.println(l+" "+cr);//L3
		int x=cr[0];
		int y=cr[1];
		// L4 : This is the same in L2.
		System.out.println(cr);//L4
		// L5: second element of array object..
		System.out.println(cr[1]);//L5
		int m=0x12996d6;
		// L6 : I can't explain about that one also...
		System.out.println(m);	//L6
	}
}

Have you tried running this to see what happens? Yes, I think that's what your question is really about.

It's interesting you commented all of the println() statements, so I think it's not the raw code you're asking about. Instead, you want to know why the results are what they are. Is that right? Assuming so, here's my output from the code you posted:

<blank line>
AB
[ I@de6f34 [ C@56ee8e // I added spaces
AB
B
19502806

I had to add spaces to the third line, because the output was being confused as code tags, so my original effort disappeared from that point. Now I'm frustrated, but I'll give you the short version:

My answers to your assessment:
L1: Yes, except I think both elements are being printed, not just one
L2: The values of cr[]'s elements, A and B
L3: These are references to l[] and cr[], a twist on the toString() method
L4: Of course
L5: Yes
L6: The value assigned to m in the previous line is being printed, but that value is crazy difficult. Try simplifying that line to:

int m = 0xFF;

and running that. Now do you see what's happening?

Originally Posted by Supun Pramoda

Is the char array must override with toString method??????

Arrays are implemented internally in the JVM, you can't extend or override anything about arrays.
Arrays derive from java.lang.Object but they don't override (redefine) the toString() method. So they have the toString() inherited from Object, that can return only something like "[I@41675ec4".
Stop, that's all.

Originally Posted by Supun Pramoda

why the integer array doesn't match in that concept???

Where is the difference you are saying?

If you want to see what was output by the println() method try this:

      ByteArrayOutputStream baos = new ByteArrayOutputStream();    //  Capture println() here
      PrintStream ps = new PrintStream(baos);
      System.setOut(ps);
		char cr[]=new char[3];
		int l[]=new int[3];
		System.out.println(cr);	//L1
		cr[0]='A';
		cr[1]='B';
		System.out.println(cr);		//L2
		l[0]=78;
		System.out.println(l+" "+cr);//L3
		int x=cr[0];
		int y=cr[1];
		System.out.println(cr);//L4
		System.out.println(cr[1]);//L5
		int m=0x12996d6;
		System.out.println(m);	//L6
      //  now show what was printed
      System.err.println("baos="+Arrays.toString(baos.toByteArray()));

Here are each of the outputs (end with 13, 10)
// baos=[0, 0, 0, 13, 10,
65, 66, 0, 13, 10,
91, 73, 64, 56, 55, 56, 49, 54, 100, 32, 91, 67, 64, 52, 50, 50, 101, 100, 101, 13, 10,
65, 66, 0, 13, 10,
66, 13, 10,
49, 57, 53, 48, 50, 56, 48, 54, 13, 10]

See what I did in post#11 to display what is printed. Change the code, execute it and copy the print out and paste it here with your questions.

what's the between char array & integer array???

A char uses 2 bytes, an int uses 4 bytes.

Originally Posted by Supun Pramoda

I mean when we change L2 like this:

l[0]=78;
System.out.println(l);

in that case the output is different from char array.

Yes, the output is different but it doesn't depend on the intrinsic differences between a char[] and a int[].

System.out.println(cr); invokes the println(char[]) of java.io.PrintStream. This println "knows" how to print all characters in the array.

System.out.println(l); invokes the println(Object) of java.io.PrintStream. This println doesn't know anything particular about the object. It simply invokes toString() on the object. And toString() for arrays returns something like "[I@41675ec4" (for int[]), "[J@697eb767" (for long[]), etc...