2D array pointers

I need some help with some 2D pointers and passing arrays to functions. Heres the code i got so far but i get a segmentation fault:

Code:

---

#include <stdio.h>

float avg(int **array, int row, int col)
{
        int i,j, count=0;
        float calculation = 0, average;
        for(i=0; i<row; i++)
        {
                for(j=0; j<col; j++)
                {
                        calculation += array[i][j];
                        count++;
                }
        }
        average = calculation/count;
        return average;
}

int main(void)
{
        int i, j;
        int row=3, col=3, array[row][col];
        for(i=0; i<row; i++)
        {
                for(j=0; j<col; j++)
                {
                        printf("Enter value %d,%d \n", i, j);
                        scanf("%d",&array[i][j]);
                }       
        }
        float average = avg(&array[0][0], row, col);
}

---

If array is defined as int ** (which, in your function, it is), then "array[i][j]" doesn't do what you want it to do. Namely, it will interpret your array[i] -- the number i spots away from where you started -- as a pointer, not an integer, and will then try to follow that pointer. It's unlikely that you typed in, not only a valid pointer, but the pointers where the right things are stored in memory.

In summary: int ** and int[][] are not compatible.

What would be the fix as I then tried this: (*(*(array+i)+j))
and still got the error

You have to make them match: either the function says

Code:

---

float avg(int array[][3], int row, int col)

---

or the variable says

Code:

---

int **array;

---

I posted literally in the last 24 hours how to fake a 2D array using int** in the middle of a thread with a long discussion of why and what, so search.

The function parameters should stay how i have them.
I tried this:
...
value = &array[i][j];
calculation += value;
....

Do i need to type cast value? what do i type cast to?

If the function parameters must stay, then array[row][col] CANNOT work. There is no way to take a 2D array and make it something that an int** can understand. You MUST use int** throughout.

Edit: Unless you are willing to forgo double-subscripting and calculate all the offsets yourself. (I.e., instead of saying array[2][1] you say something like array[2*row+1].) You'd have to cast your int** down to an int* first, though.

but i tried this:
calculation += (*(*(array+i)+j));

But i still get segmentation fault?

Well, yes. Because (*(*(array+i)+j)) makes no sense. Again, you're using the values you typed in via the for-loop as pointers, and then trying to follow those pointers.... there's nothing there, except the Land of Segfaults and Broken Dreams.

How do i get that bit working. I think i need to cast something but not sure

Have you read any of these replies?

Quote:

---

Originally Posted by tabstop

Edit: Unless you are willing to forgo double-subscripting and calculate all the offsets yourself. (I.e., instead of saying array[2][1] you say something like array[2*row+1].) You'd have to cast your int** down to an int* first, though.

---

If for whatever reason you are unable to change the function header and the variable itself (which I still recommend, since this is apparently supposed to be dynamic in some way), then you'll have to put the pointer in an int* variable and calculate the offsets yourself.

Thats where am confused, how do i do that?

Quote:

---

Originally Posted by taurus

Thats where am confused, how do i do that?

---

Do you understand how a 2-D array is laid out in memory?

Say I define array[3][3]. Then memory contains
array: [0][0] [0][1] [0][2] [1][0] [1][1] [1][2] [2][0] [2][1] [2][2]
in that order. Your job is, given [i][j], to find the offset (which item in the list it is).

How come i cant do this then
Calculation += *pointer

given Calculation is an int
?

You can. But you need to have the pointer first, and brother, (*(array+i)+j) ain't it.

As a test:

calculation += *(array+i);
Isint that a pointer?

But i get the error:
loop.c:25: error: invalid operands to binary +

and when i do this:
calculation += (int)(*(array+i));

Calculation doesnt seem to be getting set.

The array in your function is a pointer. The array in your main is not. In other words: show more code.

in main i have this:
float average = avg(&array[0][0], row, col);

and in the avg function i have this:

value = (int)(*(array+i));
calculation += value;

But calculation is not being set

Remember, in your function you have to do something like:

Code:

---

int *real_array = (int *)array;

---

since &array[0][0] is a pointer-to-int (not a pointer-to-pointer-to-int). If you just want to loop over all the entries, then you could use real_array[0] through real_array[row*col-1].

Ah i found the problem, its with the declaration type of calculation

xxsfdfd

calculation += value works whether or not calculation is a float. Perhaps value is 0? Print it out and see.

what about if I have two ints divided by each other like this:
average = (calculation/count);

where average is a float. It doesn't seem to give an answer, just 0.

Ah i fixed it, its %lf and not %d
thanks

Why not show the entire code?
I smell something bad. Maybe.

Code:

---

#include <stdio.h>

float avg(int **array, int row, int col)
{
        //LOOP i
                //LOOP j
                        //Take value of i,j and add to variable calculation
                        //Add 1 to count
                //ENDLOOP
        //ENDLOOP
        //average <- calculation/count
        //return average

        int i,j, count=0, value, calculation=0;
        float average=1.0;
        for(i=0; i<row; i++)
        {
                for(j=0; j<col; j++)
                {
                        value = (int)(*(array+i));
                        printf("Array Value = %d\n",value);
                        calculation += value;
                        printf("Calculation = %d\n",calculation);
                        //count++;
                }
        }
        printf("Calculation FINAL = %d, COUNT = %d\n",calculation, count);
        count = row*col;
        average = (calculation/count);
        printf("Average = %lf\n",average);
        return average;
}

int main(void)
{
        int i, j;
        int row=3, col=3, array[row][col];
        for(i=0; i<row; i++)
        {
                for(j=0; j<col; j++)
                {
                        printf("Enter value %d,%d \n", i, j);
                        scanf("%d",&array[i][j]);
                }       
        }
        float average = avg(*array, row, col);
        printf("Average = %lf\n",average);
}

---

Take a look

My intuition does not fail me...
Your array is 2D-array. A 2D-array is NOT T**.
You also dereference the array, which will give you T*, not T**.
And you also treat your ptp as a single pointer.

You should probably take int* in your function, and cast the array to int*, and then use the array[n] syntax instead of your horrible syntax you have now.

Example:

Code:

---

int main()
{
    int array[10][10];
    int* p = (int*)array; // Cast to a single pointer
    for (int i = 0; i < rows; i++)
        p[i] = ...;
    //etc
}

---

hmm i see this is confusing stuff, i get the basics but with double pointers (**) i get confused. Do you know of a good tutorial as iv searched online and still a bit confused, couldnt find a good one to explain things such as why cast and why do you need 2 stars etc etc.

thanks

It's not a difficult concept.
Pointers being variables, they too, have an address.
And the way pointers work, they have the type that point to, plus an extra "*" to indicate it's a pointer. So if you take the address of a pointer, what do you get? That's right, you get a ptp (pointer-to-pointer).
But you should understand that while arrays decays to a pointer, a 2D array cannot decay to a ptp. The simple reason for that is that the compiler cannot calculate the correct offset in a 2D array with just a T**.
The real types becomes T (*)[n], where n is the size of the outermost dimension. It's a pointer to an array (pointer to a 1D array).