Two-dimensional dynamic array pointer access

Question

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

如果我使用p作为二维动态数组，如何访问里面的元素？

Answer 1

If you can rely on your C implementation to support variable-length arrays (an optional feature), then a pretty good way would be to declare p as a pointer to (variable-length) array instead of a pointer to int :

int (*p)[n] = malloc(m * sizeof(*p));  // m rows, n columns

Then you access elements using ordinary double indexes, just as if you had declared an ordinary 2D array:

p[0][0] = 1;
p[m-1][n-1] = 42;
int q = p[2][1];

Most widely used C implementations do support VLAs, but Microsoft's is a notable exception.

Answer 2

I'd personally prefer using wohlstad's method, but you could also variably-modified types, which are an optional feature of C11, but will probably be mandated in C2x :

int (*p)[m] = malloc(n * sizeof *p);

This can now be used just like a normal 2d array with automatic storage duration.

int n = 12, m = 9;

int (*p)[m] = malloc(n * sizeof *p);
for (int i = 0; i < n; ++i)
    for (int j = 0; j < m; ++j)
        p[i][j] = i * m + j;
    
printf("%d\n", p[4][2]);

free(p);

Answer 3

I'll assume m is the number of columns, and n the number of rows (you can use n instead of m in my answer if it's the opposite).
In order to access the 2D array, you need 2 indices - let's call them x and y :
x index will be in the range 0 .. m-1 , and
y index will be in the range 0 .. n-1

You can calculate the index for your p array in the following way:

int p_idx = y * m + x

Then you can access your arrays element eg this way:

p[p_idx] = 111;   // set an element value
int a = p[p_idx]; // get an element value

Answer 4

You can't use p as a two-dimensional array. It's a single integer pointer. Two-dimensional (dynamically allocated) implies nested pointers. However, you can represent a two-dimensional array in a "flattened" form. Here's some code that might offer a helpful explanation:

#include <stdio.h>
#include <stdlib.h>

int main(){
    // Populating a 10x5 matrix
    int m = 10;
    int n = 5;
    int* p = (int*) malloc(m*n*sizeof(int));
    
    for (int i = 0; i < m; i++) {
        for (int j = 0; j < n; j++) {
            // Each row has n elements; to get the
            // "flattened" index, treating the MxN
            // matrix as row-major ordered (reading
            // left-to-right, and THEN down the rows):
            int flattened_index = (i * n) + j;

            // E.g., populate with multiplication table data
            p[flattened_index] = (i + 1) * (j + 1);

            printf("%d\t", p[flattened_index]);
        }
        printf("\n");
    }

    // Inversely, to convert a flattened index to a
    // row and column, you have to use modulus
    // arithmetic
    int flattened_index = 21;
    int row = flattened_index / n; // Rounded-down integer division
    int column = flattened_index % n; // Remainder after division
    
    printf("%d * %d = %d\n", row + 1, column + 1, p[flattened_index]);

    return 0;
}

This outputs:

1   2   3   4   5   
2   4   6   8   10  
3   6   9   12  15  
4   8   12  16  20  
5   10  15  20  25  
6   12  18  24  30  
7   14  21  28  35  
8   16  24  32  40  
9   18  27  36  45  
10  20  30  40  50  
5 * 2 = 10

Answer 5

You are actually creating a single-dimensional array. But still, we can use it to hold a matrix considering the fact in C a multidimensional array, eg int mat[m][n], is stored in contiguous memory itself.

#include <iostream>

int main()
{
   int m, n;

   std::cin >> m >> n;

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

   if (mat_ptr)
   {
      for (int row = 0; row < m; ++row)
      {
        for (int col = 0; col < n; ++col)
        {
            *(mat_ptr + ((row * n) + col)) = (row * n) + col;
        }
      }

      for (int row = 0; row < m; ++row)
      {
        for (int col = 0; col < n; ++col)
        {
            std::cout << *(mat_ptr + ((row * n) + col)) << " ";
        }

        std::cout << std::endl;
      }
  }

  return 0;
}