How to pass a pointer variable as a reference parameter?

Question

When you have a function like this, the formal parameter is a reference, it becomes another name for the actual argument, so that when we modify the formal parameter inside the function, the original variable outside the function is changed.

void add_five(int& a)
{
    a += 5;
}

int main()
{
    int number = 3;
    add_five(number);
    std::cout << number << std::endl;  // prints 8

    return 0;
}

I have some code which works on a linked lists. And I am passing two Node*s into the function.

void LinkedList::move_five_nodes(Node* ptr1, Node* ptr2) { ... }

...

void LinkedList::anotherFunction()
{
    Node* leader;
    Node* trailer;
    ...
    move_five_nodes(leader, trailer);
    ...
}

I think that the rvalues memory addreses inside the leader and trailer pointer variables will be assigned into the Node* lvalues ptr1 and ptr2.

Node* ptr1 = leader;
Node* ptr2 = trailer;

The issue is that ptr1 and ptr2 are independent local variables inside the function. Initially, they point to the same place as the actual argument pointers. However, my function moves some nodes, and at the end of the function, the values of ptr1 and ptr2 are changed. I want these changes to also be in the original variables leader and trailer, just like references. So even when ptr1 and ptr2 expire, leader and trailer should go into their positions.

How would I do this? Maybe type cast the pointers into int&? Or maybe use pointers to pointers? I want to pass a pointer by reference, but I'm not sure how to do that.

It sounds like you just need to redesign your function move_five_nodes. It's a little hard to tell, however. Pointers to pointers would probably work just fine. I don't understand what the issue is to using those. — OrdoFlammae
– OrdoFlammae, Commented Sep 15, 2018 at 23:43

user4290866 · Accepted Answer · 2018-09-15 23:45:51Z

7

I want to pass a pointer by reference, but I'm not sure how to do that.

For this, consider the following snippet

#include <iostream>

void test(int*& t)
{
    t = nullptr;
}

int main()
{
    int* i = new int(4);

    test(i);

    if (i == nullptr)
        std::cout << "I was passed by reference" << std::endl;
}

in which is is passed by reference to test, where it is set to nullptr and the program prints: I was passed by reference.

I think this example should make clear how to pass a pointer by reference to a function.

So in your case the function signiture must change to

void LinkedList::move_five_nodes(Node*& ptr1, Node*& ptr2) { ... }

answered Sep 15, 2018 at 23:45

user4290866

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5 Comments

Galaxy Over a year ago

One more question: int* i = new int[4]; creates a new dynamic array of 4 ints and assigns the pointer to i, while int* i = new int(4); creates a single dynamic integer initialized with 4, and assigns the pointer to i. Am I right? Do I understand this one correctly?

user4290866 Over a year ago

Yes, you are right about that. Was just an example to get a pointer.

Galaxy Over a year ago

Also nice pun i was passed by reference.

user4290866 Over a year ago

@Galaxy Wasn't intended, but i agrees ;)

kingsjester Over a year ago

Probably better with delete t; before t = nullptr to avoid a memory-leak..

Sen · Accepted Answer · 2022-10-07 04:16:11Z

1

#include <iostream>
void test(int& ref);
int main()
{
    int* pointer = new int(10);               
    std::cout << "before" << *pointer << std::endl ;
    test(*pointer);
    std::cout << "after" << *pointer << std::endl;
    delete pointer;
}

void test(int& ref)
{    
    ref = 20;
}

answered Oct 7, 2022 at 4:16

Sen

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