Access 2D array from another function in C

Question

I can access array using pointer notion in the function where the array is declared. But, I can not access from another function which throws an error: indirection requires pointer operand ('int' invalid). The code is:

#include <stdio.h>

const int ROW = 3;
const int COL = 3;

void printArray(int *ptr);

int main(void)
{
    int arr[ROW][COL];
    int count = 2;

    // initialize 2D array
    for (int i = 0; i < ROW; i++)
    {
        for (int j = 0; j < COL; j++)
        {
            arr[i][j] = count++;
        }
    }

    // print 2D array
    for (int i = 0; i < ROW; i++)
    {
        for (int j = 0; j < COL; j++)
        {
            // printf("%i\t", arr[i][j]); // using array
            printf("%i\t", *(*(arr + i) + j)); // using pointer

        }
        printf("\n");
    }

    printf("\n---------printing using function------------\n");
    printArray(&arr[0][0]);
}

void printArray(int *ptr)
{
    for (int i = 0; i < ROW; i++)
    {
        for (int j = 0; j < COL; j++)
        {
            printf("%i\t", *(*(ptr + i) + j)); // not working as expected
        }
        printf("\n");
    }
}

How can i solve this problem? any idea?

Answer 1

For starters pay attention to that in C these declarations

const int ROW = 3;
const int COL = 3;

do not declare integer constants. So your program deals with a variable length array.

If you want to define integer constants you could write for example

enum { ROW = 3, COL = 3 };

Nevertheless either declare and define the function the following way

void printArray( int ( *ptr )[COL], int row );

and

void printArray( int ( *ptr )[COL], int row )
{
    for (int i = 0; i < row; i++)
    {
        for (int j = 0; j < COL; j++)
        {
            printf("%i\t", *(*(ptr + i) + j));
        }
        printf("\n");
    }
}

and call it like

printArray( arr, ROW );

Or if to use your declaration then the function definition will look like

void printArray(int *ptr)
{
    for (int i = 0; i < ROW; i++)
    {
        for (int j = 0; j < COL; j++)
        {
            printf("%i\t", *( ptr + i * COL + j ) );
        }
        printf("\n");
    }
}

Answer 2

*(*(ptr + i) + j) de-refereneces the pointer with two levels of indirection, which isn't possible since it only got one, being an int*.

In general, refrain from crazy pointer arithmetic expressions and use readable array indexing instead: ptr[i + j]. Though your calculation is wrong too, it should be ptr[i*COL + j]

---

Though the above is referred to as "mangled 2D arrays" and is correct but kind of archaic C. By using pointers to variable-length arrays (VLA) as done in modern C programming, we can turn the code much more readable:

#include <stdio.h>

const int ROW = 3;
const int COL = 3;

void printArray(size_t row, size_t col, int array[row][col]);

int main(void)
{
    int arr[ROW][COL];
    int count = 2;

    // initialize 2D array
    for (int i = 0; i < ROW; i++)
    {
        for (int j = 0; j < COL; j++)
        {
            arr[i][j] = count++;
        }
    }

    // print 2D array
    for (int i = 0; i < ROW; i++)
    {
        for (int j = 0; j < COL; j++)
        {
            // printf("%i\t", arr[i][j]); // using array
            printf("%i\t", *(*(arr + i) + j)); // using pointer

        }
        printf("\n");
    }

    printf("\n---------printing using function------------\n");
    printArray(ROW, COL, arr);
}

void printArray(size_t row, size_t col, int array[row][col])
{
    for (size_t i = 0; i < row; i++)
    {
        for (size_t j = 0; j < col; j++)
        {
            printf("%i\t", array[i][j]);
        }
        printf("\n");
    }
}