当我在Visual Studio 2015中运行此代码时,代码可以正常运行,但是代码在代码块中生成以下错误:分段错误(核心已转储)。我也用同样的错误在ubuntu中运行了代码。
#include <iostream>
#include <immintrin.h>
struct INFO
{
unsigned int id = 0;
__m256i temp[8];
};
int main()
{
std::cout<<"Start AVX..."<<std::endl;
int _size = 100;
INFO *info = new INFO[_size];
for (int i = 0; i<_size; i++)
{
for (int k = 0; k < 8; k++)
{
info[i].temp[k] = _mm256_setr_epi8(0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19,
20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31);
}
}
std::cout<<"End AVX."<<std::endl;
return 0;
}
答案 0 :(得分:5)
问题在于,在C ++ 17之前的new
和delete
并不尊重待分配类型的对齐方式。如果您通过此简单函数查看生成的程序集:
INFO* new_test() {
int _size = 100;
INFO *info = new INFO[_size];
return info;
}
您会看到,使用C ++ 17之前的任何内容进行编译时,会调用operator new[](unsigned long)
,而对于C ++ 17,则会调用operator new[](unsigned long, std::align_val_t)
(而32
是为第二个参数传递)。
Play around with it at godbolt。
如果您不能使用C ++ 17,则可以覆盖operator new[]
(和operator delete[]
-并且还应该覆盖operator new
和operator delete
.. ):
struct INFO {
unsigned int id = 0;
__m256i temp[8];
void* operator new[](size_t size) {
// part of C11:
return aligned_alloc(alignof(INFO), size);
}
void operator delete[](void* addr) {
free(addr); // aligned_alloc is compatible with free
}
};
如果您使用-DOVERWRITE_OPERATOR_NEW
进行编译,则这是上一个示例的一部分。
请注意,在使用std::vector
(或任何其他std
-容器)时,这不能解决对齐问题,因为您需要将对齐的分配器传递给容器(不是先前的一部分)例如)。
答案 1 :(得分:0)
I found two ways to solve this problem
The first solution How to solve the 32-byte-alignment issue for AVX load/store operations?
struct INFO
{
__m256i temp[8];
unsigned int id = 0;
};
INFO *info = static_cast<INFO*>(_mm_malloc(sizeof(INFO)*_size, 32));
_mm_free(info);
The second solution
INFO *info = new INFO[_size];
for (int i = 0; i < _size; i++)
{
INFO new_info;
for (int k = 0; k < 8; k++)
{
new_info.temp[k] = _mm256_setr_epi8(0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19,
20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31);
}
info[i] = new_info;
}