How are multi-dimensional arrays formatted in memory?

Question

In C, I know I can dynamically allocate a two-dimensional array on the heap using the following code:

int** someNumbers = malloc(arrayRows*sizeof(int*));

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Answer 1

Suppose, we have a1 and a2 defined and initialized like below (c99):

int a1[2][2] = {{142,143}, {144,145}};
int **a2 = (int* []){ (int []){242,243}, (int []){244,245} };

a1 is a homogeneous 2D array with plain continuous layout in memory and expression (int*)a1 is evaluated to a pointer to its first element:

a1 --> 142 143 144 145

a2 is initialized from a heterogeneous 2D array and is a pointer to a value of type int*, i.e. dereference expression *a2 evaluates into a value of type int*, memory layout does not have to be continuous:

a2 --> p1 p2
       ...
p1 --> 242 243
       ...
p2 --> 244 245

Despite totally different memory layout and access semantics, C-language grammar for array-access expressions looks exactly the same for both homogeneous and heterogeneous 2D array:

• expression a1[1][0] will fetch value 144 out of a1 array
• expression a2[1][0] will fetch value 244 out of a2 array

Compiler knows that the access-expression for a1 operates on type int[2][2], when the access-expression for a2 operates on type int**. The generated assembly code will follow the homogeneous or heterogeneous access semantics.

The code usually crashes at run-time when array of type int[N][M] is type-casted and then accessed as type int**, for example:

((int**)a1)[1][0]   //crash on dereference of a value of type 'int'