为什么std :: accumulation这么慢? [英] Why is std::accumulate so slow?

问题描述

我试图使用简单的for循环，std::accumulate和手动展开的for循环求和数组元素.如我所料，手动展开循环是最快的循环，但更有趣的是std :: accumulate比简单循环慢得多. 这是我的代码，我使用带有-O3标志的gcc 4.7进行了编译. Visual Studio将需要不同的rdtsc函数实现.

I'm trying to sum array elements using a simple for loop, an std::accumulate and a manualy unrolled for loop. As I expect, the manually unrolled loop is the fastest one, but more interesting is that std::accumulate is much slower than simple loop. This is my code, I compiled it with gcc 4.7 with -O3 flag. Visual Studio will need different rdtsc function implementation.

#include <iostream>
#include <algorithm>
#include <numeric>
#include <stdint.h>


using namespace std;

__inline__ uint64_t rdtsc() {
  uint64_t a, d;
  __asm__ volatile ("rdtsc" : "=a" (a), "=d" (d));
  return (d<<32) | a;
}

class mytimer
{
 public:
  mytimer() { _start_time = rdtsc(); }
  void   restart() { _start_time = rdtsc(); }
  uint64_t elapsed() const
  { return  rdtsc() - _start_time; }

 private:
  uint64_t _start_time;
}; // timer

int main()
{
    const int num_samples = 1000;
    float* samples = new float[num_samples];
    mytimer timer;
    for (int i = 0; i < num_samples; i++) {
        samples[i] = 1.f;
    }
    double result = timer.elapsed();
    std::cout << "rewrite of " << (num_samples*sizeof(float)/(1024*1024)) << " Mb takes " << result << std::endl;

    timer.restart();
    float sum = 0;
    for (int i = 0; i < num_samples; i++) {
        sum += samples[i];
    }
    result = timer.elapsed();
    std::cout << "naive:\t\t" << result << ", sum = " << sum << std::endl;

    timer.restart();
    float* end = samples + num_samples;
    sum = 0;
    for(float* i = samples; i < end; i++) {
        sum += *i;
    }
    result = timer.elapsed();
    std::cout << "pointers:\t\t" << result << ", sum = " << sum << std::endl;

    timer.restart();
    sum = 0;
    sum = std::accumulate(samples, end, 0);
    result = timer.elapsed();
    std::cout << "algorithm:\t" << result << ", sum = " << sum << std::endl;

    // With ILP
    timer.restart();
    float sum0 = 0, sum1 = 0;
    sum = 0;
    for (int i = 0; i < num_samples; i+=2) {
        sum0 += samples[i];
        sum1 += samples[i+1];
    }
    sum = sum0 + sum1;
    result = timer.elapsed();
    std::cout << "ILP:\t\t" << result << ", sum = " << sum << std::endl;
}

推荐答案

对于初学者来说，您使用std::accumulate是对整数求和. 因此，您可能要付出转换每个 浮点数加到整数之前.试试:

For starters, your use of std::accumulate is summing integers. So you're probably paying the cost of converting each of the floating point to integer before adding it. Try:

sum = std::accumulate( samples, end, 0.f );

看看这有什么用.

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