程序代码：

/**++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
  *编程练习10-8，初始化一个3 * 5的二位数组，并利用一个基于变长数组的函数把
  *该数组复制到另一个一个二位数组；还要编写一个基于变长数组的函数来显示两个
  *数组的内容（该函数应该能够处理任意的N * M数组）
  **/
#include <stdio.h>

void copy(int m, int n, const double array[m][n], double target[m][n]);
void show(int m, int n, const double *array[]);

int main(void)
{
    int m=3, n=5;
    double array[3][5] = {
        {1, 2, 3, 4, 5},
        {6, 7, 8, 9, 10},
        {11, 12, 13, 14, 15}
    };

    double target[m][n];

    copy(m, n, array, target);
    show(m, n, target);

    return 0;
}

void copy(int n, int m, const double array[m][n], double target[m][n])
{
    int i, j;

    for(i=0; i<m; i++)
        for(j=0; j<n; j++)
            target[m][n] = array[m][n];
}

void show(int m, int n, const double *array[])
{
    int i, j;

    for(i=0; i<m; i++){
        for(j=0; j<n; j++)
            printf("%f  ", array[m][n]);
        putchar('\n');
    }
    printf("显示完毕。\n");
}

程序代码：

/**++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
  *编程10-18，C99标准，变长数组。
  **/

#include <stdio.h>

#define ROWS 3
#define COLS 4

int sum2d(int rows, int cols, int array[rows][cols]);

int main(void)
{
    int i, j, rs=3, cs=10;
    int junk[ROWS][COLS] = {
        {2, 4, 6, 8},
        {3, 5, 7, 9},
        {12, 10, 8, 6}
    };
    int morejunk[ROWS-1][COLS+2] = {
        {20, 30, 40, 50, 60, 70},
        {5, 6, 7, 8, 9, 10}
    };
    int var[rs][cs];    //变长数组

    for(i=0; i<rs; i++)
        for(j=0; j<cs; j++)
            var[i][j] = i*j +j;
    printf("3 * 5的数组：\n所有元素之和为：%d。\n",
        sum2d(ROWS, COLS, junk));    //变长数组
    printf("2 * 6的数组：\n所有元素之和为：%d。\n",
        sum2d(ROWS-1, COLS+1, junk));
    printf("3 * 10的数组：\n所有元素之和为：%d。\n",
        sum2d(rs, cs, var));

    return 0;
}

int sum2d(int rows, int cols, int array[rows][cols])
{
    int total=0, i, j;

    for(i=0; i<rows; i++)
        for(j=0; j<cols; j++)
            total += array[i][j];

    return total;
}