How to dynamically allocate a 2D array in C?

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; 
} 

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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

#include<stdio.h> 
#include<stdlib.h> 
  
int main() 
{ 
    int r=3, c=4, len=0; 
    int *ptr, **arr; 
    int count = 0,i,j; 
  
    len = sizeof(int *) * r + sizeof(int) * c * r; 
    arr = (int **)malloc(len); 
  
    // ptr is now pointing to the first element in of 2D array 
    ptr = arr + r; 
  
    // for loop to point rows pointer to appropriate location in 2D array 
    for(i = 0; i < r; i++) 
        arr[i] = (ptr + 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.

How to dynamically allocate a 2D array in C?

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