# How to dynamically allocate a 2D array in C?

• Difficulty Level : Hard
• Last Updated : 24 Sep, 2021

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.

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## C

 `#include ` `#include `   `void` `main()` `{` `    ``int` `r = 3, c = 4; ``//Taking number of Rows and Columns` `    ``int` `*ptr, count = 0, i;` `    ``ptr = (``int` `*)``malloc``((r * c) * ``sizeof``(``int``)); ``//Dynamically Allocating Memory` `    ``for` `(i = 0; i < r * c; i++)` `    ``{` `        ``ptr[i] = i + 1; ``//Giving value to the pointer and simultaneously printing it.` `        ``printf``(``"%d "``, ptr[i]);` `        ``if` `((i + 1) % c == 0)` `        ``{` `            ``printf``(``"\n"``);` `        ``}` `    ``}` `    ``free``(ptr);` `}`

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 ` `#include `   `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 ` `#include `   `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` `#include`   `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.