What do two left-angle brackets “<<” mean in C#?

Question

Basically the questions in the title. I\'m looking at the MVC 2 source code:

[Flags]
public enum HttpVerbs {
    Get = 1 << 0,
    Post = 1 << 1,         


        

Answer 1

When you write

1 << n

You shift the bit combination 000000001 for n times left and thus put n into the exponent of 2:

2^n

So

1 << 10

Really is

1024

For a list of say 5 items your for will cycle 32 times.

Answer 2

<< is the left bit shift operator. If you have the number 3, which is 00000011 in binary, you can write 3 << 2, which is 00001100, or 12 in decimal.

Answer 3

"Bit shift left." 1 << 0 means "take the integer value 1 and shift its bits left by zero bits." I.e., 00000001 stays unchanged. 1 << 1 means "take the integer value 1 and shift its bits left one place." 00000001 becomes 00000010.

Answer 4

Its (<<) a bitwise left shift operator, it moves the bit values of a binary object. The left operand specifies the value to be shifted and the right operand specifies the number of positions that the bits in the value are to be shifted.

In your case if the value of list.count is 4 then loop will run till i < (1<< 4) which is 16 (00010000)

00000001 << 4 = 00010000(16)

Answer 5

I know this answer is pretty much solved, but I thought the visualization might help someone.

[Fact] public void Bit_shift_left()
{
    Assert.Equal(Convert.ToInt32("0001", 2), 1 << 0); // 1
    Assert.Equal(Convert.ToInt32("0010", 2), 1 << 1); // 2
    Assert.Equal(Convert.ToInt32("0100", 2), 1 << 2); // 4
    Assert.Equal(Convert.ToInt32("1000", 2), 1 << 3); // 8
}

Answer 6

It is called left-shift operator. Take a look at the documentation

The left-shift operator causes the bit pattern in the first operand to be shifted to the left by the number of bits specified by the second operand. Bits vacated by the shift operation are zero-filled. This is a logical shift instead of a shift-and-rotate operation.

Simple example that demonstrates the left-shift operator:

for (int i = 0; i < 10; i++)
{
    var shiftedValue = 1 << i;
    Console.WriteLine(" 1 << {0} = {1} \t Binary: {2}",i,shiftedValue,Convert.ToString(shiftedValue,2).PadLeft(10,'0'));
}

//Output:

// 1 << 0 = 1      Binary: 0000000001
// 1 << 1 = 2      Binary: 0000000010
// 1 << 2 = 4      Binary: 0000000100
// 1 << 3 = 8      Binary: 0000001000
// 1 << 4 = 16     Binary: 0000010000
// 1 << 5 = 32     Binary: 0000100000
// 1 << 6 = 64     Binary: 0001000000
// 1 << 7 = 128    Binary: 0010000000
// 1 << 8 = 256    Binary: 0100000000
// 1 << 9 = 512    Binary: 1000000000

Moving one bit to left is equivelant to multiple by two.In fact,moving bits are faster than standart multiplication.Let's take a look at an example that demonstrates this fact:

Let's say we have two methods:

static void ShiftBits(long number,int count)
{
    long value = number;
    for (int i = 0; i < count; i+=128)
    {
          for (int j = 1; j < 65; j++)
          {
              value = value << j;
          }
          for (int j = 1; j < 65; j++)
          {
               value = value >> j;
          }
    }
}

static void MultipleAndDivide(long number, int count)
{
      long value = number;
      for (int i = 0; i < count; i += 128)
      {
            for (int j = 1; j < 65; j++)
            {
                value = value * (2 * j);
            }
            for (int j = 1; j < 65; j++)
            {
                value = value / (2 * j);
            }
      }
}

And we want to test them like this:

ShiftBits(1, 10000000);
ShiftBits(1, 100000000);
ShiftBits(1, 1000000000);
...
MultipleAndDivide(1, 10000000);
MultipleAndDivide(1, 100000000);
MultipleAndDivide(1, 1000000000);
...

Here is the results:

Bit manipulation 10.000.000 times: 58 milliseconds
Bit manipulation 100.000.000 times: 375 milliseconds
Bit manipulation 1.000.000.000 times: 4073 milliseconds

Multiplication and Division 10.000.000 times: 81 milliseconds
Multiplication and Division 100.000.000 times: 824 milliseconds
Multiplication and Division 1.000.000.000 times: 8224 milliseconds