Question

以下是我认为是事实的事情：

快速排序应该是相当缓存友好的。
64字节的缓存行可以包含16个32位整数或8个64位整数。

假设：

对32位整数的向量进行排序应该比排序a更快向量的64位整数。

但是当我运行下面的代码时，我得到了结果：

i16 = 7.5168 
i32 = 7.3762 
i64 = 7.5758

为什么我没有得到我想要的结果？

C ++：

#include <iostream> 
#include <vector>
#include <cstdint>
#include <algorithm>
#include <chrono>


int main() {
  const int vlength = 100'000'000;
  const int maxI = 50'000;

  std::vector<int16_t> v16;
  for (int i = 0; i < vlength; ++i) {
    v16.push_back(int16_t(i%maxI));
  }
  std::random_shuffle(std::begin(v16), std::end(v16));
  std::vector<int32_t> v32;
  std::vector<int64_t> v64;
  for (int i = 0; i < vlength; ++i) {
    v32.push_back(int32_t(v16[i]));
    v64.push_back(int64_t(v16[i]));
  }

  auto t1 = std::chrono::high_resolution_clock::now();
  std::sort(std::begin(v16), std::end(v16));
  auto t2 = std::chrono::high_resolution_clock::now();
  std :: cout << "i16 = " << (std::chrono::duration_cast<std::chrono::duration<double>>(t2 - t1)).count() << std :: endl;

  t1 = std::chrono::high_resolution_clock::now();
  std::sort(std::begin(v32), std::end(v32));
  t2 = std::chrono::high_resolution_clock::now();
  std :: cout << "i32 = " << (std::chrono::duration_cast<std::chrono::duration<double>>(t2 - t1)).count() << std :: endl;

  t1 = std::chrono::high_resolution_clock::now();
  std::sort(std::begin(v64), std::end(v64));
  t2 = std::chrono::high_resolution_clock::now();
  std :: cout << "i64 = " << (std::chrono::duration_cast<std::chrono::duration<double>>(t2 - t1)).count() << std :: endl;

}

修改为了避免缓存友好排序的问题，我还尝试了以下代码：

template <typename T>
inline void function_speed(T& vec) {
  for (auto& i : vec) {
    ++i;
  }
}

int main() {
  const int nIter = 1000;

  std::vector<int16_t> v16(1000000);
  std::vector<int32_t> v32(1000000);
  std::vector<int64_t> v64(1000000);



  auto t1 = std::chrono::high_resolution_clock::now();
  for (int i = 0; i < nIter; ++i) {
    function_speed(v16);
  }
  auto t2 = std::chrono::high_resolution_clock::now();
  std :: cout << "i16 = " << (std::chrono::duration_cast<std::chrono::duration<double>>(t2 - t1)).count()/double(nIter) << std :: endl;

  t1 = std::chrono::high_resolution_clock::now();
  for (int i = 0; i < nIter; ++i) {
    function_speed(v32);
  }
  t2 = std::chrono::high_resolution_clock::now();
  std :: cout << "i32 = " << (std::chrono::duration_cast<std::chrono::duration<double>>(t2 - t1)).count()/double(nIter) << std :: endl;

  t1 = std::chrono::high_resolution_clock::now();
  for (int i = 0; i < nIter; ++i) {
    function_speed(v64);
  }
  t2 = std::chrono::high_resolution_clock::now();
  std :: cout << "i64 = " << (std::chrono::duration_cast<std::chrono::duration<double>>(t2 - t1)).count()/double(nIter) << std :: endl;

}

典型结果：

i16 = 0.00618648
i32 = 0.00617911
i64 = 0.00606275

我知道适当的基准测试本身就是一门科学，也许我做错了。

EDIT2： 通过避免溢出，我现在开始得到更有趣的结果：

template <typename T>
inline void function_speed(T& vec) {
  for (auto& i : vec) {
    ++i;
    i %= 1000;
  }
}

给出如下结果：

i16 = 0.0143789
i32 = 0.00958941
i64 = 0.019691

如果我改为：

template <typename T>
inline void function_speed(T& vec) {
  for (auto& i : vec) {
    i = (i+1)%1000;
  }
}

我明白了：

i16 = 0.00939448
i32 = 0.00913768
i64 = 0.019615

Answer 1

错误的假设;对于绝大多数N来说，所有O（N log N）排序算法都对缓存不友好！可能的输入。

Furtehrmore，我认为优化编译器可以直接删除各种类型，而未经优化的构建当然对基准测试毫无意义。

排序32位整数与排序64位整数一样快

1 个答案: