# Bitwise operator left shifting

posted 7 years ago

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

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

Rancher

Posts: 3742

16

posted 7 years ago

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 ]

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 ]

Joanne

posted 7 years ago

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

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

Olly

Campbell Ritchie

Sheriff

Posts: 48930

58

Campbell Ritchie

Sheriff

Posts: 48930

58

posted 7 years ago

And please find the code button and indent your code; it makes it much easier to read.