SSE 错误 - 使用 m128i_i32 定义 __m128i 变量的字段 [英] SSE Error - Using m128i_i32 to define fields of a __m128i variable

问题描述

以这种方式定义 __m128i 变量后:

Upon defining a __m128i variable in this manner:

__m128i a;
a.m128i_i32[0] = 65000;

我收到以下错误:

错误:请求'a'中的成员'm128i_i32'，这是非类输入‘__m128i {aka __vector(2) long long int}’ a.m128i_i32[0] =65000;

error: request for member ‘m128i_i32’ in ‘a’, which is of non-class type ‘__m128i {aka __vector(2) long long int}’ a.m128i_i32[0] = 65000;

我已经包含了以下头文件:

I have included the followinf header files:

#include <x86intrin.h>
#include <emmintrin.h>
#include <smmintrin.h>

推荐答案

您的代码将在 Visual 下工作，其中 __m128 定义为

Your code will work under Visual where __m128 is defined as

typedef union __declspec(intrin_type) __declspec(align(16)) __m128i {
__int8              m128i_i8[16];
__int16             m128i_i16[8];
__int32             m128i_i32[4];
__int64             m128i_i64[2];
unsigned __int8     m128i_u8[16];
unsigned __int16    m128i_u16[8];
unsigned __int32    m128i_u32[4];
unsigned __int64    m128i_u64[2];
} __m128i;

所以你可以访问m128_i32，但是在g++下__m128被定义为

so you can access m128_i32, but under g++ __m128 is defined as

typedef long long __m128i __attribute__ ((__vector_size__ (16), __may_alias__));

你的代码不会被编译.

您可以通过

int32_t* p = (int32_t*)&a;
p[0] = 65000;

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