How to dynamically allocate a 2D array in C?

3.6

Following are different ways to create a 2D array on heap (or dynamically allocate a 2D array).

In the following examples, we have considered ‘r‘ as number of rows, ‘c‘ as number of columns and we created a 2D array with r = 3, c = 4 and following values

  1  2  3  4
  5  6  7  8
  9  10 11 12 

1) Using a single pointer:
A simple way is to allocate memory block of size r*c and access elements using simple pointer arithmetic.

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

int main()
{
    int r = 3, c = 4;
    int *arr = (int *)malloc(r * c * sizeof(int));

    int i, j, count = 0;
    for (i = 0; i <  r; i++)
      for (j = 0; j < c; j++)
         *(arr + i*c + j) = ++count;

    for (i = 0; i <  r; i++)
      for (j = 0; j < c; j++)
         printf("%d ", *(arr + i*c + j));

   /* Code for further processing and free the 
      dynamically allocated memory */
  
   return 0;
}

Output:

1 2 3 4 5 6 7 8 9 10 11 12

2) Using an array of pointers
We can create an array of pointers of size r. Note that from C99, C language allows variable sized arrays. After creating an array of pointers, we can dynamically allocate memory for every row.

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

int main()
{
    int r = 3, c = 4, i, j, count;

    int *arr[r];
    for (i=0; i<r; i++)
         arr[i] = (int *)malloc(c * sizeof(int));

    // Note that arr[i][j] is same as *(*(arr+i)+j)
    count = 0;
    for (i = 0; i <  r; i++)
      for (j = 0; j < c; j++)
         arr[i][j] = ++count; // Or *(*(arr+i)+j) = ++count

    for (i = 0; i <  r; i++)
      for (j = 0; j < c; j++)
         printf("%d ", arr[i][j]);

    /* Code for further processing and free the 
      dynamically allocated memory */

   return 0;
}

Output:

1 2 3 4 5 6 7 8 9 10 11 12

3) Using pointer to a pointer
We can create an array of pointers also dynamically using a double pointer. Once we have an array pointers allocated dynamically, we can dynamically allocate memory and for every row like method 2.

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

int main()
{
    int r = 3, c = 4, i, j, count;

    int **arr = (int **)malloc(r * sizeof(int *));
    for (i=0; i<r; i++)
         arr[i] = (int *)malloc(c * sizeof(int));

    // Note that arr[i][j] is same as *(*(arr+i)+j)
    count = 0;
    for (i = 0; i <  r; i++)
      for (j = 0; j < c; j++)
         arr[i][j] = ++count;  // OR *(*(arr+i)+j) = ++count

    for (i = 0; i <  r; i++)
      for (j = 0; j < c; j++)
         printf("%d ", arr[i][j]);

   /* Code for further processing and free the 
      dynamically allocated memory */

   return 0;
}

Output:

1 2 3 4 5 6 7 8 9 10 11 12

4) Using double pointer and one malloc call for all rows

#include<stdio.h>
#include<stdlib.h>
 
int main()
{
    int r=3, c=4;
    int **arr;
    int count = 0,i,j;
 
    arr  = (int **)malloc(sizeof(int *) * r);
    arr[0] = (int *)malloc(sizeof(int) * c * r);

    for(i = 0; i < r; i++)
        arr[i] = (*arr + c * i);
 
    for (i = 0; i < r; i++)
        for (j = 0; j < c; j++)
            arr[i][j] = ++count;  // OR *(*(arr+i)+j) = ++count
 
    for (i = 0; i <  r; i++)
        for (j = 0; j < c; j++)
            printf("%d ", arr[i][j]);
 
    return 0;
}

Output:

1 2 3 4 5 6 7 8 9 10 11 12

Thanks to Trishansh Bhardwaj for suggesting this 4th method.

This article is contributed by Abhay Rathi. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above

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