_mm_roti_epi8
Visual Studio 2010 SP1 is required.
Microsoft Specific
Generates the XOP instruction vprotb to rotate each of the bytes in its first source by the amount specified in the second.
__m128i _mm_roti_epi8 (
__m128i src,
int count
);
Parameters
[in] src
A 128-bit parameter that contains sixteen 8-bit unsigned integers.[in] count
An integer rotation count, preferably constant.
Return value
A 128-bit result r that contains sixteen 8-bit unsigned integers.
r[i] := (count > 0) ? rotate_left(src[i], count) :
rotate_right(src[i], -count);
Requirements
Intrinsic |
Architecture |
|---|---|
_mm_roti_epi8 |
XOP |
Header file <intrin.h>
Remarks
Each 8-bit unsigned integer value in src is rotated by the number of bits specified in count, and the 8-bit unsigned integer result is stored as the corresponding value in the destination. If the value in count is positive, the rotation is to the left (toward the most significant bit); otherwise, it is to the right.
The vprotb instruction has one form for constant arguments, another for non-constant arguments. If the value of count cannot be determined to be constant at compile time, the compiler will generate extra code to set up and use the non-constant version of vprotb. The constant version of vprotb is faster.
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.
Example
#include <stdio.h>
#include <intrin.h>
int main()
{
__m128i a, d;
int i;
for (i = 0; i < 16; i++) {
a.m128i_u8[i] = (i << 4) | (15 - i);
}
d = _mm_roti_epi8(a, -3);
printf_s("data: ");
for (i = 0; i < 16; i++) printf_s(" %02x", a.m128i_u8[i]);
printf_s("\nrotated by -3 gives ");
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 -3 gives e1 c3 a5 87 69 4b 2d 0f f0 d2 b4 96 78 5a 3c 1e