This week's book giveaway is in the OO, Patterns, UML and Refactoring forum. We're giving away four copies of Refactoring for Software Design Smells: Managing Technical Debt and have Girish Suryanarayana, Ganesh Samarthyam & Tushar Sharma on-line! See this thread for details.

char v1 = '('; int i = ~v1; System.out.println(i);

Why does the above print a negative value (-41) when the answer we get is of the form 0xffffffd6 (the actual characters may be different), after changing the 1s to 0s and vice-versa.

Can someone please explain the basic steps involved in the ~ operation. For example, int x = 9; x = ~ x;

Step 1: Convert 9 to binary: 1001 Step 2: Since the signed bit is one, the answer should be a negative integer Step 3: Now convert 0's to 1 and one's to 0. That is, 0110 Step 4: The answer: -6. This is wrong. The answer is -10. What am I doing wrong here? Please explain. Thanks

Now we have a bit pattern that we must convert to decimal. The high bit is 1 so it's negative. So let's take the two's complement and then afterwards call the number negative again.

Step 3> To get the final answer do 2's complement on the result obtained in step 2 and add the negative sign.

Hope this helps you.

Always represent number in full 32 bit integer form This is very important when you are trying to play around with bit operators.

Cheers,<br />Jay<br /> <br />(SCJP 1.4)<br />Heights of great men were not achieved in one day, they were toiling day and night while their companions slept.

Barry Gaunt
Ranch Hand

Joined: Aug 03, 2002
Posts: 7729

posted

0

In fact a bit of algebra "proves" it: Start with a positive integer x.

-x == ~x + 1 // the negative is found by flipping the bits (~x) and adding 1

Subtracting 1 from both sides:

-x - 1 == ~x

That is: ~x == -(x + 1)

So for example: ~9 == -(9 + 1) == -10.

Exercise: does it work if x < 0? for Integer.MAX_VALUE? for Integer.MIN_VALUE? [ December 06, 2004: Message edited by: Barry Gaunt ]

Sumithab Baskaran
Ranch Hand

Joined: Oct 29, 2004
Posts: 52

posted

0

Thanks Barry for the short cut method. ~x = -(x+1). It works for x <0, x= Integer.MAX_VALUE, Integer.MIN_VALUE.

So, this is all I need to know for ~x. Or do I definitely need to do it the long way - converting to one's compliment and then adding 1 to the result? Let me know so that I can move on. Thanks once again.

Jay Pawar
Ranch Hand

Joined: Aug 27, 2004
Posts: 411

posted

0

Sumitabh, For Integer.MIN_VALUE , the formula doesn't hold good. However, for ~Integer.MAX_VALUE will give you Integer.MIN_VALUE as per the formula.

Originally posted by Jay Pawar: Sumitabh, For Integer.MIN_VALUE , the formula doesn't hold good. However, for ~Integer.MAX_VALUE will give you Integer.MIN_VALUE as per the formula.