What does a bitwise shift (left or right) do and w

问题:

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 what they actually do and what some practical uses of them are.

EDIT

If the shifts are like x * 2 and x / 2, what is the real difference from actually using the * and / operators? Is there a performance difference?

回答1:

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 left. A special case is the leading 1. It indicates often 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 afaik, and maybe more) a triple-sign-operator:

1100 1100 >>> 1
0110 0110 

There is no equivalent in the other direction because it makes no 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.

回答2:

Left bit shifting to multiply by any power of two and Right bit shifting to divide by any power of two. For example x = x * 2; can also be written as x<<1 or x = x*8 can be written as x<<3 (since 2 to the power of 3 is 8). Similarly x = x / 2; is x>>1 and so on.

回答3:

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

回答4:

Left shift : It is equal to product of value which has to be shifted and 2 raised to the power of Number of Bits to be shifted.

Example :

1<<3
0000 0001  ---> 1
Shift by 1 bit
0000 0010 ----> 2 which is equal to 1*2^1
Shift By 2 bits
0000 0100 ----> 4 which is equal to 1*2^2
Shift by 3 bits
0000 1000 ----> 8 which is equal to 1*2^3

Right shift : It is equal to quotient of value which has to be shifted by 2 raised to the power of Number of Bits to be shifted.

Example :

8>>3
0000 1000  ---> 8 which is equal to 8/2^0
Shift by 1 bit
0000 0100 ----> 4 which is equal to 8/2^1
Shift By 2 bits
0000 0010 ----> 2 which is equal to 8/2^2
Shift by 3 bits
0000 0001 ----> 1 which is equal to 8/2^3

回答5:

The bit shift operators are more efficient as compared to / or * operator. In computer architecture, divide(/) or multiply(*) take more than 1 time unit and register to compute result, while, bit shift operator, is just one one register and one time unit computation.

回答6:

Left bit shifting to multiply by any power of two. Right bit shifting to divide by any power of two.

x = x << 5; // Left shift
y = y >> 5; // Right shift

In C/C++ it can be written as,

#include <math.h>

x = x * pow(2, 5);
y = y / pow(2, 5);

回答7:

Some examples:

• Bit operations for example converting to and from base64 (which is 6 bits instead of 8)
• doing power of 2 operations (1 << 4 equal to 2^4 i.e. 16)
• Writing more readable code when working with bits. For example, defining constants using 1 << 4 or 1 << 5 is more readable.

回答8:

Yes, I think performance wise you might find a difference as bitwise left and right shift operations can be performed with a complexity of o(1) with huge data set.

For eg Calculating power of 2 ^ n:-

int value = 1;
while (exponent<n)
    {
       //print out current power of 2
        value =value *2; // equivalent machine level left shift bit wise operation
        exponent++;
         }
    }

Similar code with bitwise left shift operation would be like:

value = 1 << n;

Moreover, performing bit-wise operation is like exacting replica of user level mathematical operations (which is the final machine level instructions processed by the micro controller and processor).

回答9:

Here is an example:

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

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