为什么将 unique_ptr 的 std::move() 转换为 void 函数以重置指针比调用显式 unique_ptr.reset() 花费的时间更少

Question

我制作了这些 PoC 来找出释放 unique_ptr 内存的最快方法：

#include <iostream>
#include <ctime>
#include <ratio>
#include <chrono>
#include <cstdlib>
#include <memory>

void freePointer(std::unique_ptr<int> aPointer){
    // The unique ptr memory is released here
}

int main ()
{
  using namespace std::chrono;

  // POC1
  std::unique_ptr<int> pointer (new int(3));
  high_resolution_clock::time_point t1 = high_resolution_clock::now();
  pointer.reset();
  high_resolution_clock::time_point t2 = high_resolution_clock::now();
  duration<double> time_span = duration_cast<duration<double>>(t2 - t1);
  std::cout << "POC1: It took me " << time_span.count() << " seconds." << std::endl;

  // POC2
  std::unique_ptr<int> pointer2 (new int(3));
  high_resolution_clock::time_point t3 = high_resolution_clock::now();
  freePointer(std::move(pointer2));
  high_resolution_clock::time_point t4 = high_resolution_clock::now();
  duration<double> time_span2 = duration_cast<duration<double>>(t4 - t3);
  std::cout << "POC2: It took me " << time_span2.count() << " seconds." << std::endl;
  return 0;
}

输出是：

POC1：我花了 2.573e-06 秒。
POC2：我花了 5.07e-07 秒。

PoX 之间有一个数量级的差异，我不明白原因。

有人能给我一盏灯吗？

Answer 1

#include <iostream>
#include <ctime>
#include <ratio>
#include <chrono>
#include <cstdlib>
#include <memory>
#include <vector>

void freePointer(std::unique_ptr<int> aPointer){
    // The unique ptr memory is released here
}

int main ()
{
  using namespace std::chrono;

  // POC1
  std::vector<std::unique_ptr<int>> pointers;
  for(int i=0;i<10000;++i) {
      pointers.emplace_back(new int(3));
  }
  high_resolution_clock::time_point t1 = high_resolution_clock::now();
  for(int i=0;i<10000;++i) {
      pointers[i].reset();
  }
  high_resolution_clock::time_point t2 = high_resolution_clock::now();
  duration<double> time_span = duration_cast<duration<double>>(t2 - t1);
  std::cout << "POC1: It took me " << time_span.count() << " seconds." << std::endl;
  
  // POC2
  std::vector<std::unique_ptr<int>> pointers2;
  for(int i=0;i<10000;++i) {
      pointers2.emplace_back(new int(3));
  }
  high_resolution_clock::time_point t3 = high_resolution_clock::now();
  for(int i=0;i<10000;++i) {
        freePointer(std::move(pointers2[i]));
  }
  high_resolution_clock::time_point t4 = high_resolution_clock::now();
  duration<double> time_span2 = duration_cast<duration<double>>(t4 - t3);
  std::cout << "POC2: It took me " << time_span2.count() << " seconds." << std::endl;
  
  return 0;
}

在在线编译器中对其进行测试后，结果显示 POC2 较慢但结果尚无定论，因为在某些情况下 POC1 较慢。

POC1：我花了 0.000578603 秒。
POC2：我花了 0.00146163 秒。

对于单次运行，您替换顺序，首先是 POC2，然后是 POC1，第二个将是最慢的。