2D array pointer

I got the following:

Code:

---

#include <stdio.h>
void print(int *array);
int main(void)
{
        int array[5][5] = {{1,2,3,4,5}, {1,2,3,4,5}};
        print(&array[0][0]);               
}

void print(int *array)
{
        int i, j;
        for(i=0; i<=4; i++)
        {
                for(j=0; j<=4; j++)
                {
                        printf("%d\n",array[i][j]);
                }
        }
}

---

Whats wrong with it? i get the error:

Code:

---

stringArray.c: In function `print':
stringArray.c:16: error: subscripted value is neither array nor pointer

---

That is because you are passing an array as a pointer to the first element of the string.

Now, the C language will translate arrays to pointers in itself, but the way that indexing of multidimensional arrays work, you will have to provide the dimensions of all but the first dimension.

--
Mats

so like this:
print(&array[][0]);

But that doesnt work?

Quote:

---

Originally Posted by taurus

so like this:
print(&array[][0]);

---

Erm, no, that will not compile. What I'm saying is that if you want to use a 2D array in a function, that function needs to know the size of all dimensions except the first one, so that it can multiply the sizes of the different dimensions to form the linear offset in a 1D fashion (there is no such thing as 2D or 3D memory in computers, so ALL array indices need to be made into a linear 1D reference).

In this case, you need to declare your print function with:

Code:

---

void print(int array[][5]);

---

or

Code:

---

void print(int array[5][5]);

---

The call should be made as:

Code:

---

  print(array);

---

--
Mats

But i want to play around with pointers so how would that work the same way i used int *array?

That gets tricky. You need to pass the size in X and Y dimension. Something like this:

Code:

---

print(int *arr, int xsize, int ysize);

---

And then calculate the actua y, x coordinates from that.

--
Mats

How do i calculate the offset?

Quote:

---

Originally Posted by taurus

How do i calculate the offset?

---

Think a bit about that one...

--
Mats

ah ok i found a way:
*array
then increment that

Quote:

---

Originally Posted by taurus

ah ok i found a way:
*array
then increment that

---

Erm, well, that's not going to allow you a random access to the array, but sure. What if you have a 10000 x 10000 x 10000 array, and you need to find location [500][3800][8716]? You don't really want to traverse the array one element at a time to find it, do you?

--
Mats

yea thats true.
then you could multiply those 3 numbers (addition) and say *(array + addition)
and that would work yeah? is that what you were thinking?

Quote:

---

Originally Posted by taurus

yea thats true.
then you could multiply those 3 numbers (addition) and say *(array + addition)
and that would work yeah? is that what you were thinking?

---

No.
Remembering that the rightmost number changes fastest, and to keep things small using a 5x5 array, how would things be laid out in memory? Where is array element [4][3]? Array element [2][3]? Can you spot a pattern?

so element[4][3] would be at row*4 + 3?

Right.

Ok. Check this code:

Code:

---

#include <stdlib.h>
#include <stdio.h>

void print(int** array)
{
        int i, j;
        for(i=0; i<4; i++) {
                for(j=0; j<4; j++)
                        printf("%d\t",array[i][j]);
                printf("\n");
        }
}

int main()
{
        int i,j;
        int array[4][4];
        for(i=0; i<4; i++)
                for(j=0; j<4; j++)
                        array[i][j] = 123;
        int** ptr_array;
        int* outer_ptr_array[4];
        for (i = 0; i < 4; ++i)
                outer_ptr_array[i] = array[i];
        ptr_array = outer_ptr_array;
        print(ptr_array);
}

---

So you see that you can assign a pointer to a 2D array indirectly as above. That is in case you have a function that wants int** and you don't want to change it.

Another option is to use a different type:
int (*)[5]

Like
void print(int (*array)[5], ...)

It will then work. But this is the same as typing
void print(int array[][5], ...)

Different syntax, but means the same thing. Both are pointers.