如何使用g ++对我的循环进行矢量化？

Question

我在搜索时找到的介绍性链接:

1. 6.59.14循环特定的Pragma
2. 2.100 Pragma Loop_Optimize
3. 如何给出关于循环计数的gcc的提示
4. 告诉gcc专门展开一个循环
5. 如何在C++中强制进行矢量化

正如你所看到的,大多数是C语言,但我认为它们也适用于C++.这是我的代码:

template<typename T>
//__attribute__((optimize("unroll-loops")))
//__attribute__ ((pure))
void foo(std::vector<T> &p1, size_t start,
            size_t end, const std::vector<T> &p2) {
  typename std::vector<T>::const_iterator it2 = p2.begin();
  //#pragma simd
  //#pragma omp parallel for
  //#pragma GCC ivdep Unroll Vector
  for (size_t i = start; i < end; ++i, ++it2) {
    p1[i] = p1[i] - *it2;
    p1[i] += 1;
  }
}

int main()
{
    size_t n;
    double x,y;
    n = 12800000;
    vector<double> v,u;
    for(size_t i=0; i<n; ++i) {
        x = i;
        y = i - 1;
        v.push_back(x);
        u.push_back(y);
    }
    using namespace std::chrono;

    high_resolution_clock::time_point t1 = high_resolution_clock::now();
    foo(v,0,n,u);
    high_resolution_clock::time_point t2 = high_resolution_clock::now();

    duration<double> time_span = duration_cast<duration<double>>(t2 - t1);

    std::cout << "It took me " << time_span.count() << " seconds.";
    std::cout << std::endl;
    return 0;
}

我使用了上面提到的提示,但是我没有得到任何加速,因为示例输出显示(第一次运行已取消注释#pragma GCC ivdep Unroll Vector:

samaras@samaras-A15:~/Downloads$ g++ test.cpp -O3 -std=c++0x -funroll-loops -ftree-vectorize -o test
samaras@samaras-A15:~/Downloads$ ./test
It took me 0.026575 seconds.
samaras@samaras-A15:~/Downloads$ g++ test.cpp -O3 -std=c++0x -o test
samaras@samaras-A15:~/Downloads$ ./test
It took me 0.0252697 seconds.

有什么希望吗？或者优化标志O3只是诀窍？欢迎任何加速此代码(foo函数)的建议!

我的g ++版本:

samaras@samaras-A15:~/Downloads$ g++ --version
g++ (Ubuntu 4.8.1-2ubuntu1~12.04) 4.8.1

---

请注意,循环体是随机的.我以其他形式重写它并不感兴趣.

---

编辑

答案说没有什么可以做的也是可以接受的!

Answer 1

该O3标志自动打开-ftree-vectorize.https://gcc.gnu.org/onlinedocs/gcc/Optimize-Options.html

-O3打开-O2指定的所有优化,并打开-finline-functions,-funswitch-loops,-fpredictive-commoning,-fgcse-after-reload,-ftree-loop-vectorize,-ftree-loop-distribute -patterns,-ftree-slp-vectorize,-fvect-cost-model,-ftree-partial-pre和-fipa-cp-clone选项

因此,在这两种情况下,编译器都在尝试进行循环向量化.

使用g ++ 4.8.2编译:

g++ test.cpp -O2 -std=c++0x -funroll-loops -ftree-vectorize -ftree-vectorizer-verbose=1 -o test

给出这个:

Analyzing loop at test.cpp:16                                                                                                                                                                                                                                               


Vectorizing loop at test.cpp:16                                                                                                                                                                                                                                             

test.cpp:16: note: create runtime check for data references *it2$_M_current_106 and *_39                                                                                                                                                                                    
test.cpp:16: note: created 1 versioning for alias checks.                                                                                                                                                                                                                   

test.cpp:16: note: LOOP VECTORIZED.                                                                                                                                                                                                                                         
Analyzing loop at test_old.cpp:29                                                                                                                                                                                                                                               

test.cpp:22: note: vectorized 1 loops in function.                                                                                                                                                                                                                          

test.cpp:18: note: Unroll loop 7 times                                                                                                                                                                                                                                      

test.cpp:16: note: Unroll loop 7 times                                                                                                                                                                                                                                      

test.cpp:28: note: Unroll loop 1 times  

没有-ftree-vectorize标志的编译:

g++ test.cpp -O2 -std=c++0x -funroll-loops -ftree-vectorizer-verbose=1 -o test

仅返回:

test_old.cpp:16: note: Unroll loop 7 times

test_old.cpp:28: note: Unroll loop 1 times

第16行是循环函数的开始,所以编译器肯定是向量化它.检查汇编程序也确认了这一点.

我似乎在我正在使用的笔记本电脑上获得了一些积极的缓存,这使得很难准确地测量该函数运行的时间.

但是,您可以尝试其他一些事情:

• 使用__restrict__限定符告诉编译器阵列之间没有重叠.

• 告诉编译器阵列是否对齐__builtin_assume_aligned(不可移植)

这是我生成的代码(我删除了模板,因为你想对不同的数据类型使用不同的对齐方式)

#include <iostream>
#include <chrono>
#include <vector>

void foo( double * __restrict__ p1,
          double * __restrict__ p2,
          size_t start,
          size_t end )
{
  double* pA1 = static_cast<double*>(__builtin_assume_aligned(p1, 16));
  double* pA2 = static_cast<double*>(__builtin_assume_aligned(p2, 16));

  for (size_t i = start; i < end; ++i)
  {
      pA1[i] = pA1[i] - pA2[i];
      pA1[i] += 1;
  }
}

int main()
{
    size_t n;
    double x, y;
    n = 12800000;
    std::vector<double> v,u;

    for(size_t i=0; i<n; ++i) {
        x = i;
        y = i - 1;
        v.push_back(x);
        u.push_back(y);
    }

    using namespace std::chrono;

    high_resolution_clock::time_point t1 = high_resolution_clock::now();
    foo(&v[0], &u[0], 0, n );
    high_resolution_clock::time_point t2 = high_resolution_clock::now();

    duration<double> time_span = duration_cast<duration<double>>(t2 - t1);

    std::cout << "It took me " << time_span.count() << " seconds.";
    std::cout << std::endl;

    return 0;
}

就像我说的那样,我在获得一致的时间测量方面遇到了麻烦,因此无法确认这是否会给你带来性能提升(甚至可能会降低!)