_rotl8, _rotl16
Microsoft Specific
Rotate the input values to the left to the most significant bit (MSB) by a specified number of bit positions.
unsigned char _rotl8(
unsigned char value,
unsigned char shift
);
unsigned short _rotl16(
unsigned short value,
unsigned char shift
);
Parameters
- [in] value
The value to rotate.
- [in] shift
The number of bits to rotate.
Return Value
The rotated value.
Requirements
| Intrinsic | Architecture |
|---|---|
_rotl8 |
x86, IPF, x64 |
_rotl16 |
x86, IPF, x64 |
Header file <intrin.h>
Remarks
Unlike a left-shift operation, when executing a left rotation, the high order bits that fall off the high end are moved into the least significant bit positions.
Example
// rotl.cpp
#include <stdio.h>
#include <intrin.h>
#pragma intrinsic(_rotl8, _rotl16)
int main()
{
unsigned char c = 'A', c1, c2;
for (int i = 0; i < 8; i++)
{
printf_s("Rotating 0x%x left by %d bits gives 0x%x\n", c,
i, _rotl8(c, i));
}
unsigned short s = 0x12;
int nBit = 10;
printf_s("Rotating unsigned short 0x%x left by %d bits gives 0x%x\n",
s, nBit, _rotl16(s, nBit));
}
Output
Rotating 0x41 left by 0 bits gives 0x41 Rotating 0x41 left by 1 bits gives 0x82 Rotating 0x41 left by 2 bits gives 0x5 Rotating 0x41 left by 3 bits gives 0xa Rotating 0x41 left by 4 bits gives 0x14 Rotating 0x41 left by 5 bits gives 0x28 Rotating 0x41 left by 6 bits gives 0x50 Rotating 0x41 left by 7 bits gives 0xa0 Rotating unsigned short 0x12 left by 10 bits gives 0x4800