Pointer to 2D arrays in C

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.

//defines an array of 280 pointers (1120 or 2240 bytes)
int  *pointer1 [280];

//defines a pointer (4 or 8 bytes depending on 32/64 bits platform)
int (*pointer2)[280];      //pointer to an array of 280 integers
int (*pointer3)[100][280]; //pointer to an 2D array of 100*280 integers

Using pointer2 or pointer3 produce the same binary except manipulations as ++pointer2 as pointed out by WhozCraig.

I recommend using typedef (producing same binary code as above pointer3)

typedef int myType[100][280];
myType *pointer3;

Note: Since C++11, you can also use keyword using instead of typedef

using myType = int[100][280];
myType *pointer3;

in your example:

myType *pointer;                // pointer creation
pointer = &tab1;                // assignation
(*pointer)[5][12] = 517;        // set (write)
int myint = (*pointer)[5][12];  // get (read)

Note: If the array tab1 is used within a function body => this array will be placed within the call stack memory. But the stack size is limited. Using arrays bigger than the free memory stack produces a stack overflow crash.

The full snippet is online-compilable at gcc.godbolt.org

int main()
{
    //defines an array of 280 pointers (1120 or 2240 bytes)
    int  *pointer1 [280];
    static_assert( sizeof(pointer1) == 2240, "" );

    //defines a pointer (4 or 8 bytes depending on 32/64 bits platform)
    int (*pointer2)[280];      //pointer to an array of 280 integers
    int (*pointer3)[100][280]; //pointer to an 2D array of 100*280 integers  
    static_assert( sizeof(pointer2) == 8, "" );
    static_assert( sizeof(pointer3) == 8, "" );

    // Use 'typedef' (or 'using' if you use a modern C++ compiler)
    typedef int myType[100][280];
    //using myType = int[100][280];

    int tab1[100][280];

    myType *pointer;                // pointer creation
    pointer = &tab1;                // assignation
    (*pointer)[5][12] = 517;        // set (write)
    int myint = (*pointer)[5][12];  // get (read)

    return myint;
}

int *pointer[280]; //Creates 280 pointers of type int.

In 32 bit os, 4 bytes for each pointer. so 4 * 280 = 1120 bytes.

int (*pointer)[100][280]; // Creates only one pointer which is used to point an array of [100][280] ints.

Here only 4 bytes.

Coming to your question, int (*pointer)[280]; and int (*pointer)[100][280]; are different though it points to same 2D array of [100][280].

Because if int (*pointer)[280]; is incremented, then it will points to next 1D array, but where as int (*pointer)[100][280]; crosses the whole 2D array and points to next byte. Accessing that byte may cause problem if that memory doen't belongs to your process.