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.

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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