我怎样才能在c中创建一个n维数组

Question

I was thinking of writing a function that takes n parameters and returns an n-dimensional array using those parameters as the dimensions. Now i realize an one-d and 2d array is easy to implement with pointers. For 2d array the snippet would be something like (standard way) :

int** x;
int* temp;

x = (int**)malloc(m * sizeof(int*));
temp = (int*)malloc(m*n * sizeof(int));
for (int i = 0; i < m; i++) {
  x[i] = temp + (i * n);
}

where the array is of size m*n; But the problem lies how do we find the nested loop parameters for a n-dimensional array? Is there any way to optimize the code?

Answer 1

This shows how to create an N-dimensional array and how to index its elements. These provide the basic mechanisms needed. This is something students consider when learning, but it is rarely used in practice. There are usually better ways to organize data structures. Additionally, most useful algorithms would have patterns in how they traverse the data, so it would be better to build code that updates indices efficiently in an incremental manner rather than recalculating them from scratch as shown below.

/*  Note:  For demonstration purposes only.  Depending on needs, other types
    might be used for indices and sizes, and the array type might be wrapped
    in an opaque struct rather than exposed as "int *".
*/


//  Create an array with N dimensions with sizes specified in D.
int *CreateArray(size_t N, size_t D[])
{
    //  Calculate size needed.
    size_t s = sizeof(int);
    for (size_t n = 0; n < N; ++n)
        s *= D[n];

    //  Allocate space.
    return malloc(s);
}

/*  Return a pointer to an element in an N-dimensional A array with sizes
    specified in D and indices to the particular element specified in I.
*/
int *Element(int *A, size_t N, size_t D[], size_t I[])
{
    //  Handle degenerate case.
    if (N == 0)
        return A;

    //  Map N-dimensional indices to one dimension.
    int index = I[0];
    for (size_t n = 1; n < N; ++n)
        index = index * D[n] + I[n];

    //  Return address of element.
    return &A[index];
}

Example of use:

//  Create a 3*3*7*7*9 array.
size_t Size[5] = { 3, 3, 7, 7, 9 };
int *Array = CreateArray(5, Size);

//  Set element [1][2][3][4][5] to -987.
*Element(Array, 5, Size, (size_t []) { 1, 2, 3, 4, 5 }) = -987;

Answer 2

I prefer not to use multidimensional arrays, use 1D instead:

int* pArray = (int*) malloc(m * n * sizeof(int*));

// access pArray[a][b]
int result = pArray[a*m + b];