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Bitwise operator left shifting

 
Varshini Priya
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Hi all,

Im trying to execute the below program, im not able to understand the leftshifting functionality

public class shiftdemo
{
public static void main(String[] args)
{
System.out.println("14 in binary format is:");
System.out.println("1110 \n");
System.out.println("Now right shifting the bits by 2");
System.out.println("14>>2="+(14>>2));
System.out.println("now left shifting the bits by 2");
System.out.println("14<<2="+(14<<2));
}
}
Output

C:\PROGRA~1\Java\JDK15~1.0_0\bin>java shiftdemo
14 in binary format is:
0000 00000 0000 1110

Now right shifting the bits by 2
14>>2=3
now left shifting the bits by 2
14<<2=56

Im not clear how this left shifting works. Please help me understand this. Thanks
 
Joanne Neal
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If you start with
0000 00000 0000 1110
and shift all the bits right by two places you get
0000 00000 0000 0011 (the rightmost two bits are lost)
which is 3

If you start with
0000 00000 0000 1110
and shift all the bits left by two places you get
0000 00000 0011 1000 (the two empty bits on the right are filled with zero)
which is 56
[ November 21, 2008: Message edited by: Joanne Neal ]
 
Olivier Legat
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yes what joanne says is write because each bit is worth 2^n where n is equal the bit's position from the right starting from 0.

so:
0000 00000 0000 1110 [Base-2] =
2^3 + 2^2 + 2^1 [Base-10] = 14

0000 00000 0000 0011 [Base-2] =
2^1 + 2^0 [Base-10] = 3

0000 00000 0011 1000 [Base-2] =
2^5 + 2^4 + 2^3 [Base-10] = 56

Just pointed that out in case you didn't know about binary numbers
 
Varshini Priya
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Thanks all for the detailed explanation
 
Campbell Ritchie
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Originally posted by Olivier Legat:
yes what joanne says is write because each bit is worth 2^n . . .


Except for the signed numbers (byte short int long) where the very first bit on the left is worth minus2^n.
 
Campbell Ritchie
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And please find the code button and indent your code; it makes it much easier to read.
 
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