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.

Ptr = *data; is short for *(data+0)+0 which is a pointer for first column element of the first row. the first 0 added with data is the row no., which is indirected and takes us to the first row. * (data+0) is still a address and not a value it points to (for 2D array). So, Ptr now points to the address of first column in first row. The second zero is the column no.. So, first row and first column's memory address is chosen. Using indirection (*) again would only now give value that the address holds. like * (*(data+0)+0) or **data.

Generally, if p is pointer name,i row number and j column number,

1. (*(p+i)+j) would give a memory address of a element in 2D array. i is row no. and j is col no.,
2. *(*(p+i)+j) would give the value of that element.
3. *(p+i) would access the ith row
4. to access columns, add column number to *(p+i). You may have to declare the pointer as (*p)[columns] instead of just *p. Doing so, you are declaring pointer to an 2D array.

Using pointer arithmetic is treating 2d array like 1D array. Initialize pointer *Ptr to first element (int *Ptr = *data) and then add an no. (Ptr + n) to access the columns. Adding a number higher than column number would simply continue counting the elements from first column of next row, if that exists.

It depends. It can be as simple as:

int main()
{
    int*   data[10][20]; // Fixed size known at compile time

    data[2][3] = new int(4);
}

If you want dynamic sizes at runtime you need to do some work.
But Boost has you covered:

int main()
{
   int  x;
   int  y;
   getWidthAndHeight(x,y);

   // declare a 2D array of int*
   boost::multi_array<int*,2>   data(boost::extents[x][y]);

   data[2][3] = new int(6);
}

If you are fine with jagged arrays that can grow dynamically:

int main()
{
    std::vector<std::vector<int*> >   data;

    data.push_back(std::vector<int*>(10,NULL));
    data[0][3] = new int(7);
}

Note: In all the above. I assume that the array does not own the pointer. Thus it has not been doing any management on the pointers it contains (though for brevity I have been using new int() in the examples). To do memory management correctly you need to do some more work.

To fully understand this, you must grasp the following concepts:

Arrays are not pointers!

First of all (And it's been preached enough), arrays are not pointers. Instead, in most uses, they 'decay' to the address to their first element, which can be assigned to a pointer:

int a[] = {1, 2, 3};

int *p = a; // p now points to a[0]

I assume it works this way so that the array's contents can be accessed without copying all of them. That's just a behavior of array types and is not meant to imply that they are same thing.

Multidimensional arrays

Multidimensional arrays are just a way to 'partition' memory in a way that the compiler/machine can understand and operate on.

For instance, int a[4][3][5] = an array containing 435 (60) 'chunks' of integer-sized memory.

The advantage over using int a[4][3][5] vs plain int b[60] is that they're now 'partitioned' (Easier to work with their 'chunks', if needed), and the program can now perform bound checking.

In fact, int a[4][3][5] is stored exactly like int b[60] in memory - The only difference is that the compiler now manages it as if they're separate entities of certain sizes (Specifically, four groups of three groups of five).

{
  {1, 2, 3, 4, 5}
  {6, 7, 8, 9, 10}
  {11, 12, 13, 14, 15}
}
{
  {16, 17, 18, 19, 20}
  {21, 22, 23, 24, 25}
  {26, 27, 28, 29, 30}
}
{
  {31, 32, 33, 34, 35}
  {36, 37, 38, 39, 40}
  {41, 42, 43, 44, 45}
}
{
  {46, 47, 48, 49, 50}
  {51, 52, 53, 54, 55}
  {56, 57, 58, 59, 60}
}

From this, you can clearly see that each "partition" is just an array that the program keeps track of.

Syntax

Now, arrays are syntactically different from pointers. Specifically, this means the compiler/machine will treat them differently. This may seem like a no brainer, but take a look at this:

int a[3][3];

printf("%p %p", a, a[0]);

The above example prints the same memory address twice, like this:

0x7eb5a3b4 0x7eb5a3b4

However, only one can be assigned to a pointer so directly:

int *p1 = a[0]; // RIGHT !

int *p2 = a; // WRONG !

Why can't a be assigned to a pointer but a[0] can?

This, simply, is a consequence of multidimensional arrays, and I'll explain why:

At the level of 'a', we still see that we have another 'dimension' to look forward to. At the level of 'a[0]', however, we're already in the top dimension, so as far as the program is concerned we're just looking at a normal array.

You may be asking:

Why does it matter if the array is multidimensional in regards to making a pointer for it?

It's best to think this way:

A 'decay' from a multidimensional array is not just an address, but an address with partition data (AKA it still understands that its underlying data is made of other arrays), which consists of boundaries set by the array beyond the first dimension.

This 'partition' logic cannot exist within a pointer unless we specify it:

int a[4][5][95][8];

int (*p)[5][95][8];

p = a; // p = *a[0] // p = a+0

Otherwise, the meaning of the array's sorting properties are lost.

Also note the use of parenthesis around *p: int (*p)[5][95][8] - That's to specify that we're making a pointer with these bounds, not an array of pointers with these bounds: int *p[5][95][8]

Conclusion

Let's review:

• Arrays decay to addresses if they have no other purpose in the used context
• Multidimensional arrays are just arrays of arrays - Hence, the 'decayed' address will carry the burden of "I have sub dimensions"
• Dimension data cannot exist in a pointer unless you give it to it.

In brief: multidimensional arrays decay to addresses that carry the ability to understand their contents.