What does a >> mean in Go language?

Question

I’m looking for information on Google’s Go language. In “A Tour of Go” they have this code:

const (
    Big = 1<<100
    Small = Big>>99
)

But what do << and >> mean?

You can see all of the code at http://tour.golang.org/#14

Answer 1

<< and >> are shift operators.

They work on the underlying binary representation of a number, and 'shift' the number left of the operator left or right by the amount of bits specified on the right of the operator:

1 << 1 == 2
2 << 1 == 4
111b << 3 == 111000b

Answer 2

These are bit shift left and bit shift right operators. They are the same as in the C language and it's derivatives.

x << y 

is x times 2 to the power of y

x >> y

is x divided by 2 to the power of y (fractional part discarded)

If you view the numbers as binary, than multiplication by a power of 2 shifts the bits to the left (101 * 2^3 becomes 101000) in the same way as in decimal multiplying by powers of 10 shift the number to the left (12340 * 10^3 becomes 12340000). The converse is true for division by powers of 2. It shifts the binary representation to the right. Hence the name. This is an extremely fast operation for a computer to perform by the way, so it is used a lot in performance critical bit twiddling applications like cryptography for example.

Answer 3

They are bitwise shift operators. x << y means x × 2^y, while x >> y means x × 2^−y or, equivalently, x ÷ 2^y. These operators are generally used to manipulate the binary representation of a value, where, just like with a power of 10 in decimal, multiplying or dividing by a power of two has the effect of “shifting” the digits left or right, respectively:

// Left shift:

  13 *  2 ==    26 // decimal
1101 * 10 == 11010 // binary (13 is 8 + 4 + 0 + 1)

// Right shift (brackets denote discarded portion):

  13 /  2 ==   6[.5] // decimal
1101 / 10 == 110[.1] // binary

Because you are operating on integers and a right shift typically results in fractional values, there are a couple of ways to handle how the result of a right shift is rounded. In Go, right shift is a logical shift on unsigned values and an arithmetic shift on signed values. Logical shift always rounds toward zero, while arithmetic shift always rounds down, that is, toward −∞.

Answer 4

From the specification:

Arithmetic operators

...

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

and a bit below:

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. 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.