_rotr8, _rotr16
Microsoft Specific
Rotate the input values to the right to the least significant bit (LSB) by a specified number of bit positions.
Syntax
unsigned char _rotr8(
unsigned char value,
unsigned char shift
);
unsigned short _rotr16(
unsigned short value,
unsigned char shift
);
Parameters
value
[in] The value to rotate.
shift
[in] The number of bits to rotate.
Return value
The rotated value.
Requirements
| Intrinsic | Architecture |
|---|---|
_rotr8 |
x86, ARM, x64, ARM64 |
_rotr16 |
x86, ARM, x64, ARM64 |
Header file <intrin.h>
Remarks
Unlike a right-shift operation, when executing a right rotation, the low-order bits that fall off the low end are moved into the high-order bit positions.
Example
// rotr.cpp
#include <stdio.h>
#include <intrin.h>
#pragma intrinsic(_rotr8, _rotr16)
int main()
{
unsigned char c = 'A', c1, c2;
for (int i = 0; i < 8; i++)
{
printf_s("Rotating 0x%x right by %d bits gives 0x%x\n", c,
i, _rotr8(c, i));
}
unsigned short s = 0x12;
int nBit = 10;
printf_s("Rotating unsigned short 0x%x right by %d bits "
"gives 0x%x\n",
s, nBit, _rotr16(s, nBit));
}
Rotating 0x41 right by 0 bits gives 0x41
Rotating 0x41 right by 1 bits gives 0xa0
Rotating 0x41 right by 2 bits gives 0x50
Rotating 0x41 right by 3 bits gives 0x28
Rotating 0x41 right by 4 bits gives 0x14
Rotating 0x41 right by 5 bits gives 0xa
Rotating 0x41 right by 6 bits gives 0x5
Rotating 0x41 right by 7 bits gives 0x82
Rotating unsigned short 0x12 right by 10 bits gives 0x480
END Microsoft Specific