我正在尝试将这个for循环向量化。在使用Rpass标志后,它会输出以下提示:
int someOuterVariable = 0;
for (unsigned int i = 7; i != -1; i--)
{
array[someOuterVariable + i] -= 0.3 * anotherArray[i];
}
Remark:
The cost-model indicates that vectorization is not beneficial
the cost-model indicates that interleaving is not beneficial
我希望了解这是什么意思。“交错不利”是否意味着数组索引不正确?