_rotr8, _rotr16

Artikel
08/03/2021

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

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