• 栈内矩阵相乘 v.s. 堆内矩阵相乘


    1. 栈内数组相乘

    可以定义局域变量:三个 500 x 500 的数组,共占据 3 x 500 x 500 x 8 = 6 x 10^6 bytes,差不多 6 M,栈空间约为 不到 8 M。然后进行矩阵相乘,计时得到耗时。

    2. 通过栈内指针 new 出堆内存,进行矩阵相乘

    定义 double **a, **b, **c, new 出三个 500 x 500 的数组,即大约 3 x 500 个栈内指针,指向 3 x 500 x 500 x 8 < 6M 的堆内内存,进行矩阵相乘,计时得到耗时。

    代码如下:

    #include<iostream>
    using namespace std;
    
    #include<cmath>
    #include<time.h>
    
    int n=500;
    
    double stack_multiply(void){
    
        double a[n][n], b[n][n], c[n][n];
        for(int i=0;i<n;i++){
            for(int j=0;j<n;j++){
                a[i][j] = i*j;
                b[i][j] = i*j;
            }
        }
        for(int i=0;i<n;i++){
            for(int j=0;j<n;j++){
                double y=0;
                for(int k=0;k<n;k++){
                    y += a[i][k] * b[k][j];
                }
                c[i][j] = y;
            }
        }
        return c[200][200];
    }
    
    void heap_multiply(double **a, double **b, double **c){
    
        for(int i=0;i<n;i++){
            for(int j=0;j<n;j++){
                double y=0;
                for(int k=0;k<n;k++){
                    y += a[i][k] * b[k][j];
                }
                c[i][j] = y;
            }
        }
    }
    
    int main(){
    
        clock_t t_start = clock();
        int repeat = 1E0;
        double y;
        for(int i=0;i<repeat;i++)
            y = stack_multiply();
        cout<<"		c[200][200]="<<y<<endl;
        clock_t t_end = clock();
        cout<<" It took me "<< (double)(t_end- t_start)/repeat/CLOCKS_PER_SEC<<"s to do the matrix multiplication in stack."<<endl;
    
        double **a=new double *[n];
        for(int i=0;i<n;i++){
            a[i] = new double [n];
            for(int j=0;j<n;j++){
                a[i][j] = i*j;
            }
        }
    
        int piece = 1E4;
        /*
        double ***fragment = new double ** [piece];
        for(int i=0;i<piece;i++){
            fragment[i] = new double * [piece];
            for(int j=0;j<piece;j++){
                fragment[i][j] = new double [piece];
            }
        }
        */
        double **b=new double *[n];
        for(int i=0;i<n;i++){
            b[i] = new double [n];
            for(int j=0;j<n;j++){
                b[i][j] = i*j;
            }
        }
    
        double **c = new double * [n];
        for(int i=0;i<n;i++)
            c[i] = new double [n];
            
    
        t_start = clock();
        heap_multiply(a, b, c);
        t_end = clock();
    
        cout<<"		c[200][200]="<<c[200][200]<<endl;
    
        for(int i=0;i<n;i++)
            delete [] a[i];
        delete [] a;
        
        /*
        for(int i=0;i<piece;i++){
            for(int j=0;j<piece;j++){
                delete []fragment[i][j];
            }
            delete [] fragment[i];
        }
        delete [] fragment;
        */
        
        for(int i=0;i<n;i++)
            delete [] b[i];
        delete [] b;
        for(int i=0;i<n;i++)
            delete [] c[i];
        delete [] c;
    
        cout<<" It took me "<< (double)(t_end- t_start)/CLOCKS_PER_SEC<<"s to do the matrix multiplication in heap."<<endl;
    
        return 0;
    }

    其中的 fragment 是模拟堆内存碎片化的。

    注掉 fragment 这部分以后(堆上没有碎片化,矩阵 a 与矩阵 b 紧挨着),两种矩阵相乘耗时差不多,堆上的还稍微快一点,

    g++ main.cpp

    ./a.out

        c[200][200]=1.66167e+12

     It took me 0.71875s to do the matrix multiplication in stack

        c[200][200]=1.66167e+12

     It took me 0.671875s to do the matrix multiplication in heap

    注意到,如果编译加上 -O2,耗时变为原来的 1/3 多一些,

    g++ main.cpp -O2

    ./a.out

        c[200][200]=1.66167e+12

     It took me 0.328125s to do the matrix multiplication in stack

        c[200][200]=1.66167e+12

     It took me 0.25s to do the matrix multiplication in heap

    如果用 fragment 模拟堆碎片,设置 piece = 1E1,得到

    g++ main.cpp -O2

    ./a.out

        c[200][200]=1.66167e+12

     It took me 0.296875s to do the matrix multiplication in stack

        c[200][200]=1.66167e+12

     It took me 0.234375s to do the matrix multiplication in heap

    设置 piece = 1E2 也差不多:

    g++ main.cpp -O2

    ./a.out

        c[200][200]=1.66167e+12

     It took me 0.28125s to do the matrix multiplication in stack

        c[200][200]=1.66167e+12

     It took me 0.203125s to do the matrix multiplication in heap

    所以两种方式似乎差不多。这可能是因为,从堆中读内存时,会一次得到该单元附近一块单元的值,所以只要不是离散地跳来跳去取值,这一点都会使程序变得较快。

    但在测试过程中,有一次,我得到,piece = 1E2 时,第二种变得非常慢。所以似乎也不完全确定。

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  • 原文地址:https://www.cnblogs.com/luyi07/p/10503283.html
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