Two-dimensional dynamic array pointer access - arrays

Question

p = (int *)malloc(m * n * sizeof(int));
If I use p as a two-dimensional dynamic array, how do I access the elements inside?

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 e.g. 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, e.g 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;
}