How to dynamically allocate a 2D array in C?

A 2D array can be dynamically allocated in C using a single pointer. This means that a memory block of size row*column*dataTypeSize is allocated using malloc and pointer arithmetic can be used to access the matrix elements.

A program that demonstrates this is given as follows.

Example

#include <stdio.h>
#include <stdlib.h> 
int main() {
   int row = 2, col = 3;
   int *arr = (int *)malloc(row * col * sizeof(int)); 
   int i, j;
   for (i = 0; i < row; i++)
      for (j = 0; j < col; j++)
         *(arr + i*col + j) = i + j;    
   printf("The matrix elements are:\n");
   for (i = 0; i < row; i++) {
      for (j = 0; j < col; j++) {
         printf("%d ", *(arr + i*col + j)); 
      }
      printf("\n");
   }
   free(arr); 
   return 0;
}

The output of the above program is as follows.

The matrix elements are:
0 1 2 
1 2 3

Now let us understand the above program.

The 2-D array arr is dynamically allocated using malloc. Then the 2-D array is initialized using a nested for loop and pointer arithmetic. The code snippet that shows this is as follows.

int row = 2, col = 3; 
int *arr = (int *)malloc(row * col * sizeof(int));
int i, j;
for (i = 0; i < row; i++)
   for (j = 0; j < col; j++)
      *(arr + i*col + j) = i + j; 

Then the values of the 2-D array are displayed. Finally the dynamically allocated memory is freed using free. The code snippet that shows this is as follows.

printf("The matrix elements are:\n");
for (i = 0; i < row; i++) {
   for (j = 0; j < col; j++) {
      printf("%d ", *(arr + i*col + j)); 
   }
   printf("\n");
}
free(arr);