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.

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.

In main(), the variable "array" is declared as

char array[50][50];

This is a 2500 byte piece of data. When main()'s "array" is passed about, it is a pointer to the beginning of that data. It is a pointer to a char expected to be organized in rows of 50.

Yet in function printarray(), you declare

 char **array

"array" here is a pointer to a char *pointer.

@Lucus suggestion of void printarray( char array[][50], int SIZE ) works, except that it is not generic in that your SIZE parameter must be 50.

Idea: defeat (yeech) the type of parameter array in printarray()

void printarray(void *array, int SIZE ){
    int i;
    int j;
    char *charArray = (char *) array;

    for( j = 0; j < SIZE; j++ ){
        for( i = 0; i < SIZE; i ++){
            printf( "%c ", charArray[j*SIZE + i] );
        }
        printf( "\n" );
    }
}

A more elegant solution is to make the "array" in main() an array of pointers.

// Your original printarray()
void printarray(char **array, int SIZE ){
    int i;
    int j;
    for( j = 0; j < SIZE; j++ ){
        for( i = 0; i < SIZE; i ++){
            printf( "%c ", array[j][i] );
        }
        printf( "\n" );
    }
}

// main()
char **array;
int SIZE;
// Initialization of SIZE is not shown, but let's assume SIZE = 50;
// Allocate table
array = (char **) malloc(SIZE * sizeof(char*));
  // Note: cleaner alternative syntax
  // array = malloc(sizeof *array * SIZE);
// Allocate rows
for (int row = 0; row<SIZE; row++) {
  // Note: sizeof(char) is 1. (@Carl Norum)
  // Shown here to help show difference between this malloc() and the above one.
  array[row] = (char *) malloc(SIZE * sizeof(char));
    // Note: cleaner alternative syntax
    // array[row] = malloc(sizeof(**array) * SIZE);
  }
// Initialize each element.
for (int row = 0; row<SIZE; row++) {
  for (int col = 0; col<SIZE; col++) {
    array[row][col] = 'a';  // or whatever value you want
  }
}
// Print it
printarray(array, SIZE);
...

The correct records will look like

int ( *gPtr1 )[3] = group1;
int ( *gPtr2 )[3] = group2;

And these expressions

*(gPtr1 + 3)
*(gPtr2 + 3)

are trying to access the memory after the last elements of the arrays because the arrays have only three rows. That is the type of expression *(gPtr1 + 3) is int[3] and there are only three such elements in the oridinal arrays.

or you could write

int *gPtr1 = ( int * )group1;
int *gPtr2 = ( int * )group2;

In this case using the pointers the arrays are interpretated as one-dimensional arrays with 9 elements and expressions

*(gPtr1 + 3)
*(gPtr2 + 3)

will return correspondingly

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