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<< 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
*/

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

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