Go << and>> operators

Question

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

Answer 1

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.

Answer 2

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.

All the other answers give the more technical definition, but nobody laid it out really bluntly and I thought you might want that.

Answer 3

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

Answer 4

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

Answer 5

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.

Answer 6

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

GO

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

C

#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