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The bitwise left shift operator shifts the first operand the specified number of bits to the left. The first operand is a 4-byte integer to be shifted. The second operand specifies the number of bits to shift. All bits shifted out to the left are discarded. New bits coming in from the right are zeros. Let's do a simple calculation:
| Name | Decimal | Hex | Binary |
| Op1 | 179 | 0xB3 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 1 | 1 | 0 | 0 | 1 | 1 |
| Op1 << 2 | 716 | 0x2CC | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 1 | 1 | 0 | 0 | 1 | 1 | 0 | 0 |
Left shifting a number n places is the same as multiplying it by 2n. Left shifting preserves the operand's sign and thus this rule applies to negative numbers as well. Normally, you would not multiply a number by left shifting it, because it is not clear to the reader. You would probably use it only when the performance is a critical factor.
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