Hi all, I'm a new kid on the block. I've actually borrowed the study guide from kathy sierra for SCJP 1.4 (i couldn't get 1.5, i couldnt hunt down the book in the country i'm living in), so that was how I knew abt this webbie.

Question 1 Question on Shift Operators.

Taken from kathy's book :

x gets a hexadecimal value in it.. so it looks like this

but it states : When the leftmost bit is 1, the number is negative; and when it is 0, the number is positive.

Is it referring to the "1" from the hexadecimal "8"? It does not make sense, I thought the "1" was supposed to be part of the value. x holds a value of "-2147483648". What if i want a positive value of "2147483648"? I would have to change the first digit to 0. That'd make 32 bits of zeroes and the value would be 0 and not 2147483648 when printed out.

Question 2 Question on Bitwise Complement operator.

Taken from kathy's book :

Result:

In bit representation, the conversion looks like this, 0000 0000 0000 0000 0000 0000 0000 0101 // this is binary for 5 so when i use the ~ operator, it becomes : 1111 1111 1111 1111 1111 1111 1111 1010 // is this -6 in binary ?

Hope i can get some feedback. I'm stuck with these mis-conceptions, without clarifying, i cannot move on to understand further.

Originally posted by Clemi Clemi: Hi all, I'm a new kid on the block. I've actually borrowed the study guide from kathy sierra for SCJP 1.4 (i couldn't get 1.5, i couldnt hunt down the book in the country i'm living in), so that was how I knew abt this webbie.

Question 1 Question on Shift Operators.

Taken from kathy's book :

x gets a hexadecimal value in it.. so it looks like this

but it states : When the leftmost bit is 1, the number is negative; and when it is 0, the number is positive.

Is it referring to the "1" from the hexadecimal "8"? It does not make sense, I thought the "1" was supposed to be part of the value. x holds a value of "-2147483648". What if i want a positive value of "2147483648"? I would have to change the first digit to 0. That'd make 32 bits of zeroes and the value would be 0 and not 2147483648 when printed out.

In the original number, the 1 in the left-most position does make it a negative number. But when you shift the bits left by 1, that bit falls off and makes the number non-negative.

Question 2 Question on Bitwise Complement operator.

Taken from kathy's book :

Result:

In bit representation, the conversion looks like this, 0000 0000 0000 0000 0000 0000 0000 0101 // this is binary for 5 so when i use the ~ operator, it becomes : 1111 1111 1111 1111 1111 1111 1111 1010 // is this -6 in binary ?

Hope i can get some feedback. I'm stuck with these mis-conceptions, without clarifying, i cannot move on to understand further.

Thank you!

When you have a number with a 1 in the left-most bit position, the number is negative.

To figure out what the number is, flip all of the bits and add 1 to it.

If you flip the bits back, you get the bit representation of 5. When you add 1, you get 6.

So the number is -6. [ September 02, 2007: Message edited by: Keith Lynn ]

1111 1111 1111 1111 1111 1111 1111 0000 // i'll do this

1) If the left most bit is "1", meaning it's a negative, i'll have to flip all the bits 2) After flipping, i get 0000 0000 0000 0000 0000 0000 0000 1111 3) Which represents 1 + 2 + 4 + 8 = 15 4) Thereafter, I shall add 1 to it, resulting in 16 5) Then making it a negative number, -16 (correct answer?)

Another qn : If the left most bit is "0", meaning it's a Positive, i DO NOT have to flip all the bits, is that correct?

Thanks for your first contribution to this forum and...

Welcome to the Ranch!

Hope you'll enjoy.

Only one small issue: The Java Ranch follows a certain policy regarding user names. The main reasons why and a link how to change yours you'll find here: http://www.javaranch.com/name.jsp

So, could you please change your user name before your next posting? It will not affect anything you've already posted here. Just your user name will update.

I'm posting this because I am one of the moderators of this forum.

Yours, Bu.

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