Question

I have seen this in some book/ tutorial.

When you pass in the head pointer (of linked list) into a function, you need to pass it as a double pointer.

For eg:
// This is to reverse a linked list where head points to first node.

void nReverse(digit **head)
{
    digit *prev=NULL;
    digit *curr=*head;
    digit *next;

    while(curr!=NULL)
    {
        next=curr->next;
        curr->next=prev;
        prev=curr;
        curr=next;
    }
    *head=prev;
    return;
}

This works fine.

It also works when I use single pointer like,

void nReverse(digit *head)
{
    digit *prev=NULL;
    digit *curr=head;
    digit *next;

    while(curr!=NULL)
    {
        next=curr->next;
        curr->next=prev;
        prev=curr;
        curr=next;
    }
    head=prev;
    return;
}

I tried printing the list by using the head pointer. Both the functions work fine.

Am I missing something ?

Thanks,

Answer 1

This is very C-like code, not C++.

Basically, when something is passed by-value the function operates on a copy of the data:

void foo(int i)
{
    i = 5; // copy is set to 5
}

int x = 7;
foo(x);
// x is still 7

In C, you instead pass a pointer to the variable, and can change it that way:

void foo(int* i)
{
    *i = 5; // whatever i points to is set to 5
}

int x = 7;
foo(&x);
// x is 5

For you, instead of an int it’s a digit*. (Resulting in a pointer to pointer.)

---

In C++, references were introduced. A reference is an alias to another object. So you’d do something like this:

void foo(int& i) // i is an alias to another value
{
    i = 5; // x is set to 5
}

int x = 7;
foo(x); // pass x as alias, not address of x.
// x is 5

A reference is generally preferred, since it enforces that you actually refer to an object, and simplifies both calling and operating code.

Of course in C++ you wouldn’t implement a list yourself, you’d use std::list.