Rotate Function
- Updated2025-07-30
- 3 minute(s) read
Rotates x the number of bits specified by y.
The connector pane displays the default data types for this polymorphic function.

Inputs/Outputs
y
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y specifies the number of bits in x that the function rotates. If y is greater than 0, the function shifts the first y bits at the left end of x to the right end. If y is less than 0, the function shifts the first y bits at the right end of x to the left end.
x
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If x is an 8-, 16-, 32-, or 64-bit integer, then for any value of y, y ± 8, y ± 16, y ± 32, or y ± 64 yields the same output value, respectively, as y. For example, if x is an 8-bit integer, y = 1 and y = 9 yield the same result.
x rotated left by y
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x rotated left by y is the result of the rotation. The data type of x rotated left by y is determined by the data type of the x input. |
The following table illustrates how x and y affect x rotated left by y:
| y | x | x rotated left by y | Comments |
|---|---|---|---|
| 1 | 3 (00000011) | 6 (00000110) | This function shifts the first one bit on the left, 0, to the right end. |
| 2 | 3 (00000011) | 12 (00001100) | This function shifts the first two bits on the left, 00, to the right end. |
| –3 | 3 (00000011) | 96 (01100000) | This function shifts the first three bits on the right, 011, to the left end. |
FPGA Module Details
The following details apply when you use this object in an FPGA VI.
| Single-Cycle Timed Loop | Supported. |
| Usage | N/A |
| Timing | Inside single-cycle Timed Loop--When you use this function inside a single-cycle Timed Loop, the combinatorial logic delay is proportional to the number of bits in x. Outside single-cycle Timed Loop--When you use this function outside a single-cycle Timed Loop, it takes one clock cycle and uses one register. |
| Resources | This function requires FPGA resources proportional to the number of bits in x. |
y
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x
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x rotated left by y
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