C's doesn't really treat multidimensional arrays differently from single-dimensional ones.
Arrays in C are just array of arrays
char a[2][3][4][5];
is an array 2 of array 3 of array 4 of array 5 of char.
Dereferencing/subscripting works the same for any "Array A of T":
- decay A to the address of the first element (or do nothing if you're defering/subscripting a pointer)
- add to that the index scaled by
sizeof(T)
With dereferencing/subscripting in C, when you speak of one, you speak of the other, because A[Index] or Index[A] is defined to be the same as *(A+Index) or *(Index+A).
6.5.2.1p2
A postfix expression followed by an expression in square brackets [] is a subscripted designation of an element of an array object. The definition of the subscript operator [] is that E1[E2] is identical to (*((E1)+(E2))). Because of the conversion rules that apply to the binary + operator, if E1 is an array object (equivalently, a pointer to the initial element of an array object) and E2 is an integer, E1[E2] designates the E2-th element of E1 (counting from zero).
Since in char a[2][3][4][5]; a is an array 2 of ( array 3 of array 4 of array 5 of char), a[1] would give you ((char*)&a) + 1 * sizeof(char[3][4][5]) and the result would have type char[3][4][5].
Now here's where arrays are special — arrays aren't first-class objects in C. You can't have an r-value of an array type. When you attempt to obtain one, e.g., by passing the array to a function or an operator, the array immediately decays to a pointer to its first element so the char[3][4][5] type of a[1] immediately changes to char(*)[4][5].
6.5.2.1p3
Successive subscript operators designate an element of a multidimensional array object. If E is an n-dimensional array (n >= 2) with dimensions i x j x . . . x k, then E (used as other than an lvalue) is converted to a pointer to an (n - 1)-dimensional array with dimensions j x . . . x k. If the unary * operator is applied to this pointer explicitly, or implicitly as a result of subscripting, the result is the referenced (n - 1)-dimensional array, which itself is converted into a pointer if used as other than an lvalue. It follows from this that arrays are stored in row-major order (last subscript varies fastest).
This continues recursively until you've chomped off all of the dimensions (from right to left) and are left with a real type that doesn't decay. Effectively the decay of the intermediary arrays means that intermediary derefs/subscripts don't really fetch anything — they're simply additions to a base address.
Some examples with char a[2][3][4][5];:
#include
char a[2][3][4][5];
#define ASSERT_TP(Expr,Tp) _Generic(Expr,Tp: (char*)(Expr))
int main()
{
printf("%zd\n", ASSERT_TP(a,char(*)[3][4][5])
- (char*)a); //0
printf("%zd\n", ASSERT_TP(a[1],char(*)[4][5])
- (char*)a); //60 == 1 * (3*4*5)
printf("%zd\n", ASSERT_TP(a[1][1],char(*)[5])
- (char*)a); //80 == 1 * (3*4*5) + 1 * (4*5)
}
Applied to your example:
int a[1][2] = {1,2}; // a decays to ptr to 1st element,
//i.e. to `int (*a)[2]`
printf("%p or %p\n",
*(a+0), // == a[0]; ((char*)&a) + 0*sizeof(int[2]);
// type is int[2], which decays to int*
a+0); // == a (after decay); (char*)&a + 0*sizeof(int[2]);
//type is still `int(*)[2]` (because no derefing)
Because the deref in *(a+0) didn't hit the real type yet, there was no fetch, just an addition to a base pointer
with type adjustement. Since the addition added 0, the value didn't change it remained the same as that of a decayed to a pointer to its first element (== a+0) or even &a (which would have the same numerical address but its type would be int (*)[1][2]).
