Returning a pointer to a 2d array?

**homer_3** · 03-01-2012

I'm trying to return a pointer to a 2D static array. What I've tried is

Code:

class Foo
{
public:
    void * get_array_ptr();

private:
    typedef int arr_t[10];
    arr_t foo_array[10];
}

void * Foo::get_array_ptr()
{
    return foo_array;
}

I want to use the value get_array_ptr returns just like I'd use a 2D static array, for example

Code:

class Bar : Foo
{
     void process();
}

void Bar::process()
{
    arr_t x = get_array_ptr(); //compile error here
    int* val = x[2];
    int val2 = val[3];
}

Of course, this doesn't compile. I get the error "array must be initialized with a brace enclosed initializer". Does anyone have any suggestions on how to pull this off?

**oogabooga** · 03-01-2012

You could dynamically allocate it:

Code:

class Foo {
public:
    typedef int** arr_t;
    // you may want a couple of consts defining the dimensions
    arr_t get_array_ptr();
    Foo();  // create array
    ~Foo(); // destroy array
private:
    arr_t foo_array;
};

**MK27** · 03-01-2012

Why do you want to use void* there?

Code:

#include <iostream>

using namespace std;

typedef int arr_t[10];

class Test {
		arr_t ray[3];
	public:
		Test () {}
		arr_t *getray () { return ray; }
};

int main(void) {
	Test *eg = new Test();
	arr_t *x = eg->getray();
	int n = x[2][3];
	cout << n << endl;
	return 0;
}

**homer_3** · 03-01-2012

Originally Posted by MK27

Why do you want to use void* there?

It's not the full picture. Foo actually extends another class that had get_array as a virtual function. Its return type really isn't defined. But I found a way to accomplish my goal using

Code:

typedef struct { int arr[10]; } my_struct;

and then making an array of that struct as a member var. The only hang up is I need to reference the struct's member before I do the 2nd index, which shouldn't be too big of an issue.

P.S. I can't even put a 1 liner in without code tags? Getting strict now, huh?

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