What does a bitwise shift (left or right) do and what is it used for?

Question

I\'ve seen the operators >> and << in various code that I\'ve looked at (none of which I actually understood), but I\'m just wondering

Answer 1

Left Shift

x = x * 2^value (normal operation)

x << value (bit-wise operation)

---

x = x * 16 (which is the same as 2^4)

The left shift equivalent would be x = x << 4

Right Shift

x = x / 2^value (normal arithmetic operation)

x >> value (bit-wise operation)

---

x = x / 8 (which is the same as 2^3)

The right shift equivalent would be x = x >> 3

Answer 2

Here is an applet where you can exercise some bit-operations, including shifting.

You have a collection of bits, and you move some of them beyond their bounds:

1111 1110 << 2
1111 1000

It is filled from the right with fresh zeros. :)

0001 1111 >> 3
0000 0011

Filled from the left. A special case is the leading 1. It often indicates a negative value - depending on the language and datatype. So often it is wanted, that if you shift right, the first bit stays as it is.

1100 1100 >> 1
1110 0110

And it is conserved over multiple shifts:

1100 1100 >> 2
1111 0011

If you don't want the first bit to be preserved, you use (in Java, Scala, C++, C as far as I know, and maybe more) a triple-sign-operator:

1100 1100 >>> 1
0110 0110

There isn't any equivalent in the other direction, because it doesn't make any sense - maybe in your very special context, but not in general.

Mathematically, a left-shift is a *=2, 2 left-shifts is a *=4 and so on. A right-shift is a /= 2 and so on.

Answer 3

Here is an example:

#include"stdio.h"
#include"conio.h"

void main()
{
    int rm, vivek;
    clrscr();
    printf("Enter any numbers\t(E.g., 1, 2, 5");
    scanf("%d", &rm); // rm = 5(0101) << 2 (two step add zero's), so the value is 10100
    printf("This left shift value%d=%d", rm, rm<<4);
    printf("This right shift value%d=%d", rm, rm>>2);
    getch();
}