Passing pointer of multi-dimensional pointer array to a function - c++

Question

ok so suppose I have a function myFunction. Then in main i have a multi dimensional array of pointers. I want to pass a pointer to this array of pointers into myFunction. How would I do that? I know that If you want to pass an int to my function, one can write the function as
myfunct( int x) { ...}
What would that type of x be if I have to pass a pointer to an array of pointers? Thanks in advance :D

Answer 1

Typically you want to modify the elements of an array rather then the actual pointer. The actual pointer is given by malloc and if you change it, by writing directly to the value, it won't affect the memory allocation (except you might loose the initial pointer...).
This might be what you're looking for in a 2D array.
void myfunct(int** ptr,int items, int array_items)
{
//some code
}
int main(int argc, char *argv[])
{
const auto items = 5;
const auto array_items = 7;
int** multi_dimensional_array = reinterpret_cast<int**>(std::malloc(items * sizeof(int*)));
for (auto i = 0 ;i < items;++i)
{
multi_dimensional_array[i] = static_cast<int*>(std::malloc(sizeof(int) * array_items));
}
myfunct(multi_dimensional_array,items,array_items);
//deallocate
}

Answer 2

Wrap your multidimensional array inside a class. That way you can carry the data and dimensions in one block and passing it around is as simple as moving around any other class.
Remember to observe the Rules of Three, Five, and Zero, whichever best applies to how you store your array inside your class. std::vector is a personal favourite because it allows you to use the Rule of Zero.
For example:
#include <iostream>
#include <vector>
struct unspecified
{
};
template<class TYPE>
class TwoDee{
int rows;
int cols;
std::vector<TYPE> data;
public:
TwoDee(int row, int col):rows(row), cols(col), data(rows*cols)
{
// does nothing. All of the heavy lifting was in the initializer
}
// std::vector eliminates the need for destructor, assignment operators, and copy
//and move constructors. All hail the Rule of Zero!
//add a convenience method for easy access to the vector
TYPE & operator()(size_t row, size_t col)
{
return data[row*cols+col];
}
TYPE operator()(size_t row, size_t col) const
{
return data[row*cols+col];
}
};
void function(TwoDee<unspecified *> & matrix)
{
// does stuff to matrix
}
int main()
{
TwoDee<unspecified *> test(10,10);
function(test);
}

Answer 3

To directly answer your question, typically the type passed will be int * for a vector of int, and int ** for a 2D array of int
void myfunct( int **x)
{
x[2][1] = 25;
return;
}
If for some reason you wanted that to be an array of int pointers instead of int you need an extra *.
void myfunct( int ***x)
{
*(x[2][1]) = 25;
return;
}

Answer 4

Let me first try to interpret the exact type that you want to deal with. I suppose in your main function there is a "multidimensional array" which stores pointers for each element. As an example, let's say you have a 3-dimensional array of pointer to integer type.
Assume that you know the size of the array:
C style array will look like this:
int *a[4][3][2];
that means a is a 4x3x2 array, and each element in the array is a pointer to integer. So overall you now have 24 pointers to integer in total, as can be seen by testing the result of sizeof(a) / sizeof(int*) (the result should be 24). Okay, so far so good. But now I guess what you want is a pointer to the array a mentioned above, say b, so b is defined
int *(*b)[4][3][2] = &a;
Notice that although now b looks intimidating, in the end it is just a pointer which just stores an address, and sizeof(b) / sizeof(int*) gives 1 as the result. (The * inside parenthesis indicates b is pointer type, so b is a pointer to a "multidimensional array" of pointers to integer.)
Now to pass b to myFunction, just give the same type of b as argument type in the declaration:
void myFunction(int *(*x)[4][3][2]) {
// do something
}
And that's it! You can directly use myFunction(b) to invoke this function. Also, you can test that inside myFunction, x is still of the size of one pointer, and *x is of the size of 24 pointers.
*Note that since we are passing a pointer to array type into the function, the array-to-pointer decay does not apply here.
Assume you don't know the size of the array at compile time:
Say you have int N1 = 4, N2 = 3, N3 = 2; and you want to initialize a N1xN2xN3 array of pointer to integer, you cannot directly do that on the stack.
You could initialize use new or malloc as suggested in #Mikhail's answer, but that approach takes nested loops for multidimensional arrays and you need to do nested loops again when freeing the memory. So as #user4581301 suggests, std::vector provides a good wrapper for dynamic size array, which do not need us to free the memory by ourselves. Yeah!
The desired array a can be written this way (still looks kind of ugly, but without explicit loops and bother of freeing memory)
std::vector<std::vector<std::vector<int*>>> a (N1,
std::vector<std::vector<int*>> (N2,
std::vector<int*> (N3)
)
);
Now, b (the pointer to a) can be written as
auto *b = &a;
You can now pass b with
void myFunction(std::vector<std::vector<std::vector<int*>>>* x) {
// do something
}
Notice that the * before x means x is a pointer.