Bala Arul

Hi Fei,
So, what is wrong with the above conversion method.
Bala.

Hi Paul,
1) When you have same precedence, the evaluation will be left to right. * and / are in same precedence.
2) 2895729615 = 10101100100110010101001111001111 in binary.
Note that Most Significant Bit has 1 means it is negative.
Bala.

An instance of a class is called an object. IE, Object is an instance (of a class).
Bala Arul

Hi Faisal,
1) Hex to Binary
Convert every digit into 4 bit Binary.
e.g) A6 = 1010 0110
2) Binary to Hex
Group by 4 bits from right to left, then convert each group to Hex equivalent.
e.g) 101011 = 0010 1011 = 2B
3) Oct to Bin and Bin to Oct would have the procedure as above, but with group of 3 bits instead of 4 bits.
4) For Oct to Hex and Hex to Oct, there need to be an intermediate binary conversion. i.e, Oct to Bin to Hex and Hex to Bin to Oct.
Hope that helps.
B Arul.

[This message has been edited by Bala Arul (edited May 08, 2001).]

Hi Shah,
g.fillRect(52, 53, 50, 50) will use graphics context's current color (I think that is what you mean by ForegroundColor).
Bala Arul.

Hi Tracy,

Originally posted by Tracy Qi:
I am still confused here.
array[i++] = array[6]
But I think after this operation i is still 5. so how come i = 6?

[] is evaluted first then post-increment ++. That means,
array[i++] yealds array[5] and i = 6.
After this the evaluation of the expression on the right side of the equal sign is performed, and I hope that you are clear on that.
Bala Arul.

[This message has been edited by Bala Arul (edited May 03, 2001).]

Since x is reference of type B(base class), x.x is always refer to the Base integer member x, not the hiding integer member of A.
But when it comes to methods, actual object method would be invoked. That means if A has a over-riding method, that is the one is invoked. If there is none of over-riding method, then the base class method is invoked.
Bala Arul.

Hi raimondas zemaitis,

Originally posted by raimondas zemaitis:
I would really appreciate if someone could visit my site and drop me a message if you experience the same
problems.

I just visited the site and have same probloms as you described.
B Arul

Hi,
U can access fm this URL.
http://www.geocities.com/velmurugan_p/
Bala Arul.

Hi Keith,
I hope that the following thread would help.
http://www.javaranch.com/ubb/Forum24/HTML/009073.html
Bala Arul
[This message has been edited by Bala Arul (edited April 04, 2001).]

Answers : -1, -1.
After 1 is left shifted by 31, the high order bit is 1. When this is right shifted by 31, all the 32 bits are 1's. That is -1. In case of i which is again right shifted by 1, still all the bits are 1's. Thus the result.
Bala Arul.

Hi,
As an integer -0's binary representation is as same as 0, by taking 2's compliment.
B Arul.

In my opinion, 0.0 is equals to -0.0. Sign only indicates how the zero was achieved, whether from positive or negative value.
B Arul

Hi Sunit,
It is all about operator precedence.
First a[i] is computed and printed, where 'i' is (a=b)[3].
To evaluate (a=b)[3], first assign array b into a, then take the subscript [3]. So a and b are denoting the same array { 2, 3, 1, 0 }. Thus i is equal to 0 (a[3] / b[3]). But as I mentioned before, a[i] is evaluated first. So a[0] from original array a ({ 1, 2, 3, 4 }) is printed. If you analyze array a after this 'println' statement, you will find that a referencing the array b ({ 2, 3, 1, 0 }).
Bala Arul.

Hi Pratiti,
False.
i) String s = "Hello" + "Java"; yields "HelloJava" as the value of s.
ii) String s = "Hello";
String s2= "Java";
s.concat( s2 );
After all the above 3 statements executed, s still refer to the value "Hello". s.concat(s2) does not change the s's value, but return new String object whose value is "HelloJava"
Bala.