C - multidimensional array memory allocation based on user input

Question

I have seen the following declaration of two dimensional array.

 int arr[][3] = { {1,2,3}, {4,5,6}};

My question is how can I allocate following multidimensional array in run time based on user input of first dimension?

 #define M 10
 #define N 15
 int arr[][M][N]

Answer 1

Start by declaring a pointer suitable for accessing the array:

int (*array)[M][N];

Then allocate memory for the array based on the user input:

array = malloc(P * sizeof(*array)); // P is the value obtained from the user

Then use the pointer as if it was a 3D array:

array[x][y][z] = 42;

Don't forget to free the memory when you're done with it.

• aliceangel: Can you please explain these lines... int(*array)[M][N] and malloc. In malloc, what is the sizeof returns? the size of whole array or the size of int pointer?
• user3386109: @userfirst785935 sizeof(array) is the size of a pointer, but sizeof(*array) is the size of an array of 150 ints. So if you try printf("%zu\n", sizeof(*array)), the output should be 600 (assuming an int is 4 bytes). When you multiply the 600 by the value provided by the user, the result is the number of bytes in the whole 3D array.

Answer 2

C allows variable-length arrays. So after reading the first dimension from the user, you can declare the array with that size.

int n;
printf("How big is it? ");
scanf("%d", &n);
int arr[n][M][N];

• phuclv: note that it's only allowed in C99, and optional in C11
• Stargateur: @GRC VLA allow any level of nested array.