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_mm_rot_epi8

Visual Studio 2010

Visual Studio 2010 SP1 is required.

Microsoft Specific

Generates the XOP instruction vprotb to rotate each of the bytes in its first source by an amount specified in the second.

__m128i _mm_rot_epi8 (
   __m128i src,
   __m128i counts
);

[in] src

A 128-bit parameter that contains sixteen 8-bit unsigned integers.

[in] counts

A 128-bit parameter that contains sixteen 8-bit signed integers.

A 128-bit result r that contains sixteen 8-bit unsigned integers.

r[i] := (counts[i] > 0) ? rotate_left(src[i], counts[i]) : 
                          rotate_right(src[i], -counts[i]);

Intrinsic

Architecture

_mm_rot_epi8

XOP

Header file <intrin.h>

Each 8-bit unsigned integer value in src is rotated by the number of bits specified in the corresponding value in counts, and the 8-bit unsigned integer result is stored as the corresponding value in the destination. If the value in counts is positive, the rotation is to the left (toward the most significant bit); otherwise, it is to the right.

The vprotb instruction is part of the XOP family of instructions. Before you use this intrinsic, you must ensure that the processor supports this instruction. To determine hardware support for this instruction, call the __cpuid intrinsic with InfoType = 0x80000001 and check bit 11 of CPUInfo[2] (ECX). This bit is 1 when the instruction is supported, and 0 otherwise.

#include <stdio.h>
#include <intrin.h>
int main()
{
    __m128i a, b, d;
    int i;
    for (i = 0; i < 16; i++) {
        a.m128i_u8[i] = (i << 4) | (15 - i);
        b.m128i_i8[i] = i - 8;
    }
    printf_s("data:       ");
    for (i = 0; i < 16; i++) printf_s(" %02x", a.m128i_u8[i]);
    printf_s("\nrotated by  ");
    for (i = 0; i < 16; i++) printf_s(" %2d", b.m128i_i8[i]);
    d = _mm_rot_epi8(a, b);
    printf_s("\ngives       ");
    for (i = 0; i < 16; i++) printf_s(" %02x", d.m128i_u8[i]);
    printf_s("\n");
}
data:        0f 1e 2d 3c 4b 5a 69 78 87 96 a5 b4 c3 d2 e1 f0
rotated by   -8 -7 -6 -5 -4 -3 -2 -1  0  1  2  3  4  5  6  7
gives        0f 3c b4 e1 b4 4b 5a 3c 87 2d 96 a5 3c 5a 78 78
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