Question

我想计算这两种方式的访问时间：Row major和Column major 因为我们知道C / C ++是Row major，所以当我们以第一种方式处理（Row major）时，我们应该更快。但是用C ++语言看这段代码

#include <iostream>
#include <time.h>
#include <cstdio>
clock_t RowMajor()
{
char* buf =new char [20000,20000];
clock_t start = clock();
for (int  i = 0; i < 20000; i++) 
    for (int j = 0; j <20000; j++) 
        {
         ++buf[i,j];
        } 
    clock_t elapsed = clock() - start;
   delete [] buf;
return elapsed  ;
} 
clock_t ColumnMajor()
 {
char* buf =new char[20000,20000];
clock_t start = clock();
for (int  i = 0; i < 20000; i++) 
    for (int  j = 0; j < 20000; j++) 
    {
        ++buf[j,i]; 
    }
clock_t elapsed = clock() - start;
 delete [] buf;
return elapsed ;

}

 int main()
  {     
        std::cout << "Process Started." << std::endl;
        printf( "ColumnMajor took  %lu microSeconds. \n",   ColumnMajor()*1000000/ (CLOCKS_PER_SEC)  );
        printf( "RowMajor took  %lu microSeconds. \n",  RowMajor() *1000000/ (CLOCKS_PER_SEC) );
        std::cout << "done" << std::endl; return 0; 
  }

但每当我运行此代码时，我会得到不同的答案，有时Rowmajor时间比列主要时间更重，有时相反，任何帮助都是有帮助的。

Answer 1

在c ++中，coma operator不能用于创建/访问矩阵的东西。要创建matrix，您需要跟踪with和height并将所有内存分配为数组。基本上你需要创建一个数字或元素等价于矩阵中元素数量的向量，你可以通过x + y * width得到每个元素。

clock_t RowMajor()
{
    int width = 20000;
    int height = 20000;
    char* buf = new char[width * height];
    clock_t start = clock();
    for (int j = 0; j < height; j++)
        for (int i = 0; i < width; i++)
        {
            ++buf[i + width * j];
        }
    clock_t elapsed = clock() - start;
    delete[] buf;
    return elapsed;
}

ColumnMajor需要使用buf

访问buf[j * width + i];

另一种创建矩阵的方法（来自评论，感谢James Kanze）就是创建缓冲区，如：char (*buf)[20000] = new char[20000][200000]。在这种情况下，访问缓冲区就像：buf[i][j]

最安全的方法是使用std :: vector或array，并避免使用new / delete。使用std :: vector来防止缓冲区写溢出：

clock_t RowMajor()
{
    int width = 20000;
    int height = 20000;
    std::vector<char> buf;
    buf.resize(width * height);
    clock_t start = clock();
    for (int j = 0; j <height; j++)
        for (int i = 0; i <width; i++)
        {
            ++buf[i + j * width];
        }
    clock_t elapsed = clock() - start;
    return elapsed;
}

Answer 2

感谢Raxvan，这是最终的代码到目前为止工作正常

#include <iostream>
#include <time.h>
#include <cstdio>
#include <windows.h>

int calc = 0;


clock_t RowMajor()
{
int width = 20000;
int height = 20000;
char* buf = new char[width * height];
clock_t start = clock();
for (int j = 0; j < height; j++)
    for (int i = 0; i < width; i++)
    {
        ++buf[i + width * j];
    }
clock_t elapsed = clock() - start;
delete[] buf;
return elapsed;
}
clock_t ColumnMajor()
{
 int width = 20000;
int height = 20000;
char* buf = new char[width * height];
clock_t start = clock();
for (int j = 0; j < height; j++)
    for (int i = 0; i < width; i++)
    {
        ++buf[j + width * i];
    }
clock_t elapsed = clock() - start;
delete[] buf;
return elapsed;
}




 int main()
  {     
        std::cout << "Process Started." << std::endl;
        calc= ColumnMajor() /CLOCKS_PER_SEC  ;
        printf( "ColumnMajor took  %lu  . \n",  calc );
        calc=RowMajor()/CLOCKS_PER_SEC ;
        printf( "RowMajor took  %lu  . \n", calc );
        std::cout << "done" << std::endl; return 0; 
  }

C ++计算行主要和列主要访问时间

2 个答案: