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c - gcc 不会矢量化简单循环

转载 作者:行者123 更新时间:2023-12-04 12:11:43 25 4
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我正在尝试从 gcc auto-vectorize documentation 中矢量化示例 4 的简化版本.对于我的生活,我不知道该怎么做;

typedef int aint __attribute__ ((__aligned__(16)));
void foo1 (int n, aint * restrict px, aint *restrict qx) {

/* feature: support for (aligned) pointer accesses. */
int *__restrict p = __builtin_assume_aligned (px, 16);
int *__restrict q = __builtin_assume_aligned (qx, 16);

while (n--){
//*p++ += *q++; <- this is vectorized
p[n] += q[n]; // This isn't!
}
}

我正在运行 gcc 4.7.2 gcc -o apps/craft_dbsplit.o -c -Wall -g -ggdb -O3 -msse2 -funsafe-math-optimizations -ffast-math -ftree-vectorize -ftree-vectorizer-verbose=5 -funsafe-loop-optimizations -std =c99

它回复:

Analyzing loop at apps/craft_dbsplit.c:388

388: dependence distance = 0.
388: dependence distance == 0 between *D.9363_14 and *D.9363_14
388: dependence distance = 0.
388: accesses have the same alignment.
388: dependence distance modulo vf == 0 between *D.9363_14 and *D.9363_14
388: vect_model_load_cost: unaligned supported by hardware.
388: vect_get_data_access_cost: inside_cost = 2, outside_cost = 0.
388: vect_model_store_cost: unaligned supported by hardware.
388: vect_get_data_access_cost: inside_cost = 2, outside_cost = 0.
388: Alignment of access forced using peeling.
388: Vectorizing an unaligned access.
388: vect_model_load_cost: aligned.
388: vect_model_load_cost: inside_cost = 1, outside_cost = 0 .
388: vect_model_load_cost: unaligned supported by hardware.
388: vect_model_load_cost: inside_cost = 2, outside_cost = 0 .
388: vect_model_simple_cost: inside_cost = 1, outside_cost = 0 .
388: not vectorized: relevant stmt not supported: *D.9363_14 = D.9367_20;

apps/craft_dbsplit.c:382: note: vectorized 0 loops in function.

最佳答案

循环从高地址到低地址。您的 gcc 将 vector 操作视为从低地址运行到高地址,因此没有意识到它可以矢量化。您的“优化”使循环成为 while (n--),实际上阻止了更相关的优化。尝试

#include <stddef.h>

void foo1 (size_t n, int *restrict px, int const *restrict qx)
{
int *restrict p = __builtin_assume_aligned(px, 16);
int const *restrict q = __builtin_assume_aligned(qx, 16);
size_t i = 0;
while (i < n)
{
p[i] += q[i];
i++;
}
}

关于c - gcc 不会矢量化简单循环,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/35555150/

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