Allocating 2D array without using array notation

Question

Been working on an assignment that requires creating and traversing 2D array of integers, but without using array notation, such as a[1][2] or even ( (a+i)+j). It is required to use pointer arithmetic to access each individual element.

So far my approach has been to allocate it as such. Included are comments that show what I think my code should be accomplishing, feel free to correct me otherwise.

int **a;  // 2d array, or pointer to array of pointers
row = 5;     
count = 5;

// allocate space
a = new int* [row];   // create array of pointers
int *p = *m;          // point to first array
for (int r = 0; r < row; r++)
{
    a = new int [col];  // create space within each array
    a++;
}

Row and columns are dynamic, and is passed in as parameters in the function. 5 is just used as an example here.

To fill the 5 x 5 2d array with elements (example 5), I use the following :

int *pRow = *a;            // get array of pointers
int *pElement = &pRow[0];  // get first element from the first array
int r = 0;
int c = 0;

while (r < row)
{
    while (c < col)
    {
        int element = 5;    
        *pElement = element;  // set first element to 5
        pElement++;           // go to next element
        c++;                  
    }
    c = 0;
    r++;
}

The problem is that *pElement = element throws a EXC_BAD_ACCESS error so there's something that I'm doing wrong.

Could anyone point out my mistake or correct my understanding of pointers, if it is found wanting?

--

Also, as a side note, is it correct to say,

int *pRow = *a; 
int *pElement = &pRow[0];

accesses the same thing (which is the first element) as

int *pElement = &**a[0];

Answer 1

Why don't you just declare a simple array. A 2D array can be flatten into a 1d. It would greatly simplify the arithmetic.

int* a = new int[row * column];

for (int iRow = 0; iRow < (row - 1); ++iRow)
    for (int jCol = 0; jCol < (column - 1) ; ++jCol)
{
// here you can access a[iRow * row + jCol]
}

Answer 2

*pElement = element throws a EXC_BAD_ACCESS error

You will have to reset pElement between each column, right now, it will continue up to 20 steps after the limit of the first column.

Then you have to be more careful using the & operator. It takes the address of the variable. The [0] operation takes the first value in an array, these two means that: &pRow[0] == pRow => true

And since you are not supposed to use array notation, pRow would suffice for that.

This is also true semantically, however, as I stated above, there is no need for &var[0] , it would evaluate to var anyway.

Also, as a side note, is it correct to say,

int *pRow = *a; 
int *pElement = &pRow[0];

accesses the same thing (which is the first element) as

int *pElement = &**a[0];