How do I pass an array to function by reference? - c++

Question

I want to pass array to function by reference. This function will dynamically allocate the elements.
This code give me an error message: access violation
#include <iostream>
using namespace std;
void func(int* ptr)
{
ptr = new int[2];
ptr[0] = 1;
ptr[1] = 2;
}
void main()
{
int* arr = 0;
func(arr);
cout << arr[0] << endl; // error here
}

Answer 1

C++ passes arguments by value. If you want to pass an int pointer by reference, you need to tell C++ the type is "reference to pointer to int": int*&.
Thus, your function prototype should be
void func(int*& ptr);
Note that you are still not handling pointers or references to arrays, but rather a pointer to the first element, which is close but not the same.
If you really wanted to pass an array of size 2 by reference, your function prototype would look like this:
void func(int(&my_array)[2]);

Answer 2

First of all function main shall have return type int.
As for your question then either you have to pass the pointer by reference or pass pointer to pointer. Or you could simply return pointer from the function.
Here is a demonstrative program that shows the three approaches. Take into account that you could use smart pointer std::unique_ptr.
#include <iostream>
int * func()
{
int *ptr = new int[2];
ptr[0] = 1;
ptr[1] = 2;
return ptr;
}
void func( int **ptr )
{
delete [] *ptr;
*ptr = new int[2];
( *ptr )[0] = 3;
( *ptr )[1] = 4;
}
void func( int * &ptr )
{
delete [] ptr;
ptr = new int[2];
ptr[0] = 5;
ptr[1] = 6;
}
int main()
{
int *arr = func();
std::cout << arr[0] << ", " << arr[1] << std::endl;
func( &arr );
std::cout << arr[0] << ", " << arr[1] << std::endl;
func( arr );
std::cout << arr[0] << ", " << arr[1] << std::endl;
delete [] arr;
}
And one more example of using std::unique_ptr
#include <iostream>
#include <memory>
std::unique_ptr<int[]> func()
{
std::unique_ptr<int[]> p( new int[2] );
p[0] = 1;
p[1] = 2;
return p;
}
void func( std::unique_ptr<int[]> &p )
{
p.reset( new int[2] );
p[0] = 3;
p[1] = 4;
}
int main()
{
std::unique_ptr<int[]> p = func();
std::cout << p[0] << ", " << p[1] << std::endl;
func( p );
std::cout << p[0] << ", " << p[1] << std::endl;
}
As you see in this case you need not bother about calling operator delete []