How does C allocate space for a 2D (3D…) array when using malloc?

Question

I am having a problem understanding how C allocates space for a 2D (or more dimensional) array, especially when we use malloc and the like. Take the program in

Answer 1

There are various ways of doing it, depending on how you're going to access it. You can ensure that the main body of the array is contiguous, or you can avoid that. For arrays of strings, you often don't bother with making the body of the array contiguous. For 2D (etc) arrays of integers or doubles, you usually do make the body of the array contiguous.

In the examples, the data type of the array is the generic type T, assumed to be numeric so the array elements can be assigned 0. The examples do not error check the memory allocations; they should in production code.

Array access with computed indexes — contiguous array body

int n1 = 5;
int n2 = 6;

T *a = malloc(n1 * n2 * sizeof(T));

for (int i = 0; i < n1; i++)
    for (int j = 0; j < n2; j++)
        a[i * n2 + j] = 0;

free(a);

Array access with double subscripts — contiguous array body

int n1 = 5;
int n2 = 6;

T **a = malloc(n1 * sizeof(T*));
T  *b = malloc(n1 * n2 * sizeof(T));

for (int i = 0; i < n1; i++)
    a[i] = &b[i * n2];

for (int i = 0; i < n1; i++)
    for (int j = 0; j < n2; j++)
        a[i][j] = 0;

free(b);
free(a);

Array access with double subscripts — discontiguous array body

int n1 = 5;
int n2 = 6;

T **a = malloc(n1 * sizeof(T*));

for (int i = 0; i < n1; i++)
    a[i] = malloc(n2 * sizeof(T));

for (int i = 0; i < n1; i++)
    for (int j = 0; j < n2; j++)
        a[i][j] = 0;

for (int i = 0; i < n1; i++)
    free(a[i]);
free(a);