create a pointer to a 2D array

Question

I have a bool array[size][size] and want to pass a pointer of it to a function to change its value.

void change_to_true(???, int i, int j) { // don't know what type it should be
  array[i][j] = 1;
}

How do I create a pointer of the array, and how should I pass the pointer to the function so that I can change its value? BTW it's c code.

Thanks

Edit: I tried

static bool array[size][size];
bool (*array_t)[index] = array;
void change_to_true(bool array, int i, int j);

Compiler told me subscripted value is neither array nor pointer I tried to change the type definition of the function parameter from bool array to bool* array , still not work.

I also tried bool (*array_t)[][size] = &array and bool (*array_t)[size][size] = &array; for the pointer, still not work.

Also since the size of the array is decided during the runtime, I can't have a enum or a global array defined that way.

Answer 1

这应该工作：

void change_to_true(bool** array, int i, int j)

Answer 2

Simple solution — but no pointer to 2D array

The function should be simply:

#include <stdbool.h>

enum { size = 24 };

extern void change_to_true(bool array[][size], int i, int j);

void change_to_true(bool array[][size], int i, int j)
{
    array[i][j] = true;
}

int main(void)
{
    bool array[size][size];

    for (int i = 0; i < size; i++)
    {
        for (int j = 0; j < size; j++)
        {
             array[i][j] = false;
             change_to_true(array, i, j);
        }
    }
}

I'm assuming, of course, that you have a C99 or C2011 compiler. If you're stuck with C89, then you'll have to provide definitions for bool and true ; the rest does not need to change. I'm not convinced the function call is worth it, compared with the assignment of false .

Solution using pointer to 2D array

Note that the code above does not, strictly, create a pointer to a 2D array. If you are really, really sure that's what you want, then the notation changes:

#include <stdbool.h>

enum { size = 24 };

extern void change_to_true(bool (*array)[][size], int i, int j);

void change_to_true(bool (*array)[][size], int i, int j)
{
    (*array)[i][j] = true;
}

int main(void)
{
    bool array[size][size];

    for (int i = 0; i < size; i++)
    {
        for (int j = 0; j < size; j++)
        {
             array[i][j] = false;
             change_to_true(&array, i, j);
        }
    }
}

There's no benefit to using a formal 'pointer to array' that I can think of, though — not in this context.

Solution using C99 variable length arrays

In C99, you can also use a VLA — variable length array — like this:

#include <stdbool.h>

extern void change_to_true(int size, bool array[][size], int i, int j);

void change_to_true(int size, bool array[][size], int i, int j)
{
    array[i][j] = true;
}

int main(void)
{
    for (int size = 2; size < 10; size++)
    {
        bool array[size][size];

        for (int i = 0; i < size; i++)
        {
            for (int j = 0; j < size; j++)
            {
                array[i][j] = false;
                change_to_true(size, array, i, j);
            }
        }
    }
}

Note that the size must precede the array declaration in the function declaration. And yes, you also write this with pointer to array notation (actually, I edited the pointer to array notation code first, then realized that I wanted the simpler version).

---

Which to use?

Unless you have compelling reasons not mentioned in the question, the first version is the code to use.

Answer 3

This should answer you question.

In short, you make a pointer to the first element and then pass it around.

Answer 4

[EDIT] C supports bool since C 99 Besides that, by declaring it that way, the actual array is a pointer to a memory zone with all the size^2 instances. This means that you can pass array to change it in the function, and in your header declare that you are receiving bool array[][]. That way it should work.