'Go << and >> operators

Could someone please explain to me the usage of << and >> in Go? I guess it is similar to some other languages.

Solution 1:^[1]

The super (possibly over) simplified definition is just that << is used for "times 2" and >> is for "divided by 2" - and the number after it is how many times.

So n << x is "n times 2, x times". And y >> z is "y divided by 2, z times".

For example, 1 << 5 is "1 times 2, 5 times" or 32. And 32 >> 5 is "32 divided by 2, 5 times" or 1.

Solution 2:^[2]

From the spec at http://golang.org/doc/go_spec.html, it seems that at least with integers, it's a binary shift. for example, binary 0b00001000 >> 1 would be 0b00000100, and 0b00001000 << 1 would be 0b00010000.

Go apparently doesn't accept the 0b notation for binary integers. I was just using it for the example. In decimal, 8 >> 1 is 4, and 8 << 1 is 16. Shifting left by one is the same as multiplication by 2, and shifting right by one is the same as dividing by two, discarding any remainder.

Solution 3:^[3]

The << and >> operators are Go Arithmetic Operators.

<<   left shift             integer << unsigned integer
>>   right shift            integer >> unsigned integer

The shift operators shift the left operand by the shift count specified by the right operand. They implement arithmetic shifts if the left operand is a signed integer and logical shifts if it is an unsigned integer. The shift count must be an unsigned integer. There is no upper limit on the shift count. Shifts behave as if the left operand is shifted n times by 1 for a shift count of n. As a result, x << 1 is the same as x*2 and x >> 1 is the same as x/2 but truncated towards negative infinity.

Solution 4:^[4]

They are basically Arithmetic operators and its the same in other languages here is a basic PHP , C , Go Example

package main

import (
    "fmt"
)

func main() {
    var t , i uint
    t , i = 1 , 1

    for i = 1 ; i < 10 ; i++ {
        fmt.Printf("%d << %d = %d \n", t , i , t<<i)
    }


    fmt.Println()

    t = 512
    for i = 1 ; i < 10 ; i++ {
        fmt.Printf("%d >> %d = %d \n", t , i , t>>i)
    }

}

GO Demo

#include <stdio.h>
int main()
{

    int t = 1 ;
    int i = 1 ;

    for(i = 1; i < 10; i++) {
        printf("%d << %d = %d \n", t, i, t << i);
    }

        printf("\n");

    t = 512;

    for(i = 1; i < 10; i++) {
        printf("%d >> %d = %d \n", t, i, t >> i);
    }    

  return 0;
}

C Demo

PHP

$t = $i = 1;

for($i = 1; $i < 10; $i++) {
    printf("%d << %d = %d \n", $t, $i, $t << $i);
}

print PHP_EOL;

$t = 512;

for($i = 1; $i < 10; $i++) {
    printf("%d >> %d = %d \n", $t, $i, $t >> $i);
}

PHP Demo

They would all output

1 << 1 = 2 
1 << 2 = 4 
1 << 3 = 8 
1 << 4 = 16 
1 << 5 = 32 
1 << 6 = 64 
1 << 7 = 128 
1 << 8 = 256 
1 << 9 = 512 

512 >> 1 = 256 
512 >> 2 = 128 
512 >> 3 = 64 
512 >> 4 = 32 
512 >> 5 = 16 
512 >> 6 = 8 
512 >> 7 = 4 
512 >> 8 = 2 
512 >> 9 = 1

Solution 5:^[5]

<< is left shift. >> is sign-extending right shift when the left operand is a signed integer, and is zero-extending right shift when the left operand is an unsigned integer.

To better understand >> think of

var u uint32 = 0x80000000;
var i int32 = -2;

u >> 1;  // Is 0x40000000 similar to >>> in Java
i >> 1;  // Is -1 similar to >> in Java

So when applied to an unsigned integer, the bits at the left are filled with zero, whereas when applied to a signed integer, the bits at the left are filled with the leftmost bit (which is 1 when the signed integer is negative as per 2's complement).

Solution 6:^[6]

Go's << and >> are similar to shifts (that is: division or multiplication by a power of 2) in other languages, but because Go is a safer language than C/C++ it does some extra work when the shift count is a number.

Shift instructions in x86 CPUs consider only 5 bits (6 bits on 64-bit x86 CPUs) of the shift count. In languages like C/C++, the shift operator translates into a single CPU instruction.

The following Go code

x := 10
y := uint(1025)  // A big shift count
println(x >> y)
println(x << y)

prints

0
0

while a C/C++ program would print

5
20

Solution 7:^[7]

In decimal math, when we multiply or divide by 10, we effect the zeros on the end of the number.

In binary, 2 has the same effect. So we are adding a zero to the end, or removing the last digit

Solution 8:^[8]

n << x = n * 2^x Example: 3 << 5 = 3 * 2^5 = 96

y >> z = y / 2^z Example: 512 >> 4 = 512 / 2^4 = 32

Solution 9:^[9]

These are Right bitwise and left bitwise operators

Sources

This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.

Source: Stack Overflow

Solution	Source
Solution 1	Amin Shojaei
Solution 2	peterh
Solution 3	peterSO
Solution 4	Baba
Solution 5	Mike Samuel
Solution 6
Solution 7	robert king
Solution 8	spatial
Solution 9	dÄ±lo sÃ¼rÃ¼cÃ¼

'Go << and >> operators

Solution 1:[1]

Solution 2:[2]

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Solution 5:[5]

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Solution 9:[9]