<< Operator (C# Reference)
The left-shift operator (<<) shifts its first operand left by the number of bits specified by its second operand. The type of the second operand must be an int or a type that has a predefined implicit numeric conversion to int.
If the first operand is an int or uint (32-bit quantity), the shift count is given by the low-order five bits of the second operand. That is, the actual shift count is 0 to 31 bits.
If the first operand is a long or ulong (64-bit quantity), the shift count is given by the low-order six bits of the second operand. That is, the actual shift count is 0 to 63 bits.
Any high-order bits that are not within the range of the type of the first operand after the shift are discarded, and the low-order empty bits are zero-filled. Shift operations never cause overflows.
User-defined types can overload the << operator (see operator); the type of the first operand must be the user-defined type, and the type of the second operand must be int. When a binary operator is overloaded, the corresponding assignment operator, if any, is also implicitly overloaded.
class MainClass11 { static void Main() { int i = 1; long lg = 1; // Shift i one bit to the left. The result is 2. Console.WriteLine("0x{0:x}", i << 1); // In binary, 33 is 100001. Because the value of the five low-order // bits is 1, the result of the shift is again 2. Console.WriteLine("0x{0:x}", i << 33); // Because the type of lg is long, the shift is the value of the six // low-order bits. In this example, the shift is 33, and the value of // lg is shifted 33 bits to the left. // In binary: 10 0000 0000 0000 0000 0000 0000 0000 0000 // In hexadecimal: 2 0 0 0 0 0 0 0 0 Console.WriteLine("0x{0:x}", lg << 33); } } /* Output: 0x2 0x2 0x200000000 */
