Both your examples are equivalent. However, the first one is less obvious and more "hacky", while the second one clearly states your intention.
int (*pointer)[280];
pointer = tab1;
pointer
points to an 1D array of 280 integers. In your assignment, you actually assign the first row of tab1
. This works since you can implicitly cast arrays to pointers (to the first element).
When you are using pointer[5][12]
, C treats pointer
as an array of arrays (pointer[5]
is of type int[280]
), so there is another implicit cast here (at least semantically).
In your second example, you explicitly create a pointer to a 2D array:
int (*pointer)[100][280];
pointer = &tab1;
The semantics are clearer here: *pointer
is a 2D array, so you need to access it using (*pointer)[i][j]
.
Both solutions use the same amount of memory (1 pointer) and will most likely run equally fast. Under the hood, both pointers will even point to the same memory location (the first element of the tab1
array), and it is possible that your compiler will even generate the same code.
The first solution is "more advanced" since one needs quite a deep understanding on how arrays and pointers work in C to understand what is going on. The second one is more explicit.