# << 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.

## Remarks

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.

## Example

``````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
*/
``````

## Comments

Note that i<<1 and i<<33 give the same result, because 1 and 33 have the same low-order five bits.

## See Also

C# Operators

#### Concepts

C# Programming Guide

C# Reference