Go << and >> operators
GoBitwise OperatorsGo Problem Overview
Could someone please explain to me the usage of <<
and >>
in Go? I guess it is similar to some other languages.
Go Solutions
Solution 1 - Go
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 - Go
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 - Go
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 - Go
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)
}
}
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;
}
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);
}
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 - Go
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 6 - Go
<<
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 7 - Go
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 - Go
These are Right bitwise and left bitwise operators
Solution 9 - Go
n << x = n * 2^x Example: 3 << 5 = 3 * 2^5 = 96
y >> z = y / 2^z Example: 512 >> 4 = 512 / 2^4 = 32