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How to dynamically allocate a 2D array in C?

  • Difficulty Level : Hard
  • Last Updated : 12 Jan, 2022
 

Following are different ways to create a 2D array on the 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 the following values 

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

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

C




#include <stdio.h>
#include <stdlib.h>
 
int main(void)
{
    int r = 3, c = 4;
 
    int* ptr = malloc((r * c) * sizeof(int));
 
    /* Putting 1 to 12 in the 1D array in a sequence */
    for (int i = 0; i < r * c; i++)
        ptr[i] = i + 1;
 
    /* Accessing the array values as if it was a 2D array */
    for (int i = 0; i < r; i++) {
        for (int j = 0; j < c; j++)
            printf("%d ", ptr[i * c + j]);
        printf("\n");
    }
 
    free(ptr);
 
    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.
 

C




#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 */
 
    for (int i = 0; i < r; i++)
        free(arr[i]);
 
    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. 
 

C




#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 */
 
    for (int i = 0; i < r; i++)
        free(arr[i]);
 
    free(arr);
 
    return 0;
}

Output: 

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

4) Using double pointer and one malloc call 
 

C




#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 = (int *)(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.
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

5) Using a pointer to Variable Length Array.

The dimensions of VLA are bound to the type of the variable. Therefore one form a pointer to an array with run-time defined shape.
The pointer has to be dereferenced before subscripting with syntax (*arr)[i][j].
 

C




#include <stdio.h>
#include <stdlib.h>
 
int main()
{
    int row = 3, col = 4, i, j, count;
 
    int (*arr)[row][col] = malloc(sizeof *arr);
     
    count = 0;
    for (i = 0; i < row; i++)
        for (j = 0; j < col; j++)
            (*arr)[i][j] = ++count;
 
    for (i = 0; i < row; i++)
        for (j = 0; j < col; j++)
            printf("%d ", (*arr)[i][j]);
 
    free(arr);
     
    return 0;
}

6) Using a pointer to the first row of VLA

Similar to 5 but allows arr[i][j] syntax.

C




#include <stdio.h>
#include <stdlib.h>
 
int main()
{
    int row = 3, col = 4, i, j, count;
 
    int (*arr)[col] = calloc(row, sizeof *arr);
     
    count = 0;
    for (i = 0; i < row; i++)
        for (j = 0; j < col; j++)
            arr[i][j] = ++count;
 
    for (i = 0; i < row; i++)
        for (j = 0; j < col; j++)
            printf("%d ", arr[i][j]);
 
    free(arr);
     
    return 0;
}


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