一个动态多维数组的实现

 只是一个想法的实现，投入到实际应用中还需进一步完善
作者：李嘉
未经书目许可，禁止任何商业目的的转载

#include "stdafx.h"
#include <stdio.h>
#include <stdlib.h>
#include <stdarg.h>
//////////////////////////////////////////////////////////////////////////
//
// Code By  : LiJia, http://blog.csdn.net/imjj
// Date     : 2007-05-04
// Purpose  : a mulidim, dynamic array show
//////////////////////////////////////////////////////////////////////////
struct Array
{
    int dim;
    int * dims;
    int * data;
};
Array * array_create(int dim, ...)
{
    if (dim < 1 )
        return NULL;
    int i;
    int * dims;
    int total_elems = 1;
    Array * array = NULL;
    // parse arguments
    va_list argptr;
    va_start(argptr, dim);
    dims = (int*)malloc(dim * sizeof(int));
    for (i = 0; i < dim; i++)
    {
        dims[i] = va_arg(argptr, int);
        total_elems *= dims[i];
    }
    // create the array
    array = (Array*)malloc(sizeof(Array));
    array->dim = dim;
    array->dims = dims;
    array->data = (int*)malloc(total_elems * sizeof(int));
    return array;
}
void array_free(Array * array)
{
    free(array->dims);
    free(array->data);
    free(array);
}
int * array_get(Array * array, ...)
{
    int i, j;
    int pos = 0;
    int n;
    int exp;
    va_list argptr;
    va_start(argptr, array);
    for(i = 0; i < array->dim; i++)
    {
        n = va_arg(argptr, int);
        //TODO: check n in range [0, array->dims[i]] here
        exp = 1;
        //TODO: a more effective pos calculate algorithm
        for (j = array->dim - 1; j > i; j--)
            exp *= array->dims[j];
        pos += n * exp;
    }
    return array->data + pos;
}
int array_size(Array * array)
{
    int size = 1;
    int i;
    for (i = 0; i < array->dim; i++)
        size *= array->dims[i];
    return size;
}
const int I = 5, J = 5, K = 5;
int main(int agrc, char * argv [])
{
    Array * array = array_create(3, I, J, K); // create a 3 dim and size [5][5][5] array
    int i, j, k;
    int size = array_size(array);
    for (i = 0; i < size; i++)
        array->data[i] = (i + 1);
    for(int i = 0; i < I; i++)
    {
        for (j = 0; j < J; j++)
        {
            for (k = 0; k < K; k++)
                printf("a[%d][%d][%d] = %d/n",
                        i, j, k,
                        *array_get(array, i, j , k));
        }
    }
    array_free(array);
    return 0;
}