In most of the explanations in the book the author insists of using **list passing instead of *list, however as per my understanding I feel there is nothing wrong in *list. Please someone explain me in detail if i am wrong. For example, to delete head of a linked list, the author says the below code is wrong.
void RemoveHead(node *head)
{
node *temp = head-> next; /* line 1 */
free(head);
head = temp;
}
Instead he says the below code must be used
void RemoveHead(node **head)
{
node *temp = (*head)-> next;
free(*head);
*head = temp;
}
In your first example:
head = temp;
Does nothing outside the function, it only sets the local variable head to temp inside the function. Whatever variable the caller passed in will remain unchanged. The author is correct, you need to use pointers or (better) references (if you're using C++).
Well. When you want to modify a integer variable inside a function, you pass it by reference. That is, you pass its address.
int x = 5;
void modify_value (int* x)
{
(*x) = 7; // is not 5 anymore
}
With pointers is the same. If you want to modify a pointer. You have to pass it by reference. That is, you have to pass its address. Which is a pointer to a pointer.
int* ptr_x = &x;
void modify_pointer (int** ptr_x)
{
*ptr_x = NULL; // is not &x anymore
}
Author is right. ** is effectively pass by reference (i.e. you can change 'head').
In your version (pass by value), head remains unchanged in the calling code.
When passing a node* as the argument, you can free, and modify the contents of the memory address pointed by that pointer. However, modifications performed on the pointer itself will not be seen from outside the function, since the pointer is passed by value.
On you second example, since you are passing a pointer to the head pointer, you can actually modify the actual head pointer, not just the memory pointed by it.
So in your second function, when *head = temp; is executed, you are modifying the address to which the pointer passed as argument points to.
Therefore, the author is right.
The author is correct. In your first example, you're modifying a local copy of head - the caller won't notice that anything has changed. head will still have been freed, so you're on track for a crash in pretty short order. You need to pass a pointer to modify the caller's variable.
Related
This is kind of silly, but I can't really explain why this is happening. As an exercise, I wanted to reverse a singly-linkedlist and I did this by defining the method:
class solution {
void reverseLinkedList(Node*& head) {
Node* curr = head;
Node* prev = NULL;
while (curr != NULL) {
Node* _next = curr->next;
curr->next = prev;
prev = curr;
curr = _next;
}
head = prev;
}
In my main function, I make the call
solution s;
s.reverseLinkedList(head);
Node* iterator = head;
while (iterator != NULL) {
std::cout<<iterator->data<<std::endl;
iterator = iterator->next;
}
Where I previously defined my head pointer to some linkedlist. The while loop is for printing my linkedlist and the function does it job. This only worked after I passed the head node by reference; I initially tried to pass Node* head instead of Node*& head in the beginning, and it only printed the first element of my linkedlist (and without reversing it). For example, if I didn't pass by reference for a list 1->2->3, I would print out just 1.
I thought passing a pointer would be enough? Why did I get such weird behaviour without passing by reference>
Local variables in C++ (stored in the stack) have block scope, i.e., they run out of scope after the block in which they are defined is executed.
When you are passing in a pointer to the function, a copy of the pointer is created and that copy is what is passed. Once the function is executed, the variables in the function workspace run out of scope. Any non-static Automatic variables created within the function are destroyed.
When you pass in by reference you don't pass in a copy of the variable but you pass in the actual variable, thereby any changes made to the variable are reflected on the actual variable passed into the function(by reference).
I would like to point out that the pointer to the next node is stored in memory and has an address to the location it is stored. So if you want to not pass in by reference you can do this:
Use a pointer to the pointer, which points to the memory location at which pointer variable(address) to the next node is stored
Pass in this to the function (not by reference)
Dereference the pointer and store the new address you want to point to.
I know this is a bit confusing, but look into this small piece of code that adds a node to a linked list.
void addNode(Node** head, int newData)
{
Node* newNode = new Node;
newNode->data = newData; // Can also be done using (*newNode).data
newNode->next = *head;
*head = newNode;
}
I thought passing a pointer would be enough?
void reverseLinkedList(Node* head) // pass pointer by value
// head is a copy here
Passing a pointer by value creates a copy to be used inside the function.
Any changes made to that pointer inside the function is only reflected in the function scope. Since those changes are only reflected in the pointer's copy and not in the original pointer.
Once the pointer (copy) goes out of scope, those changes are "discarded" due to end of life.
Thus, you need a reference.
void reverseLinkedList(Node&* head) // changes made in head will be
// reflected in original head pointer
When you pass a pointer regularly (IE by value) it creates a copy of the pointer. Any changes made to this pointer do not effect the original pointer.
Passing a pointer by reference is sending a reference to that pointer (very similar to passing a pointer to a pointer) and therefor any changes made to that pointer are effecting its 'original' state.
For example:
//WRONG does not modify the original pointer, causes memory-leak.
void init(Object* ptr, int sz) {
ptr = new T[sz];
}
vs
//Correct original pointer is a set to a new block of memory
void init(Object*& ptr, int sz) {
ptr = new T[sz];
}
In
void reverseLinkedList(Node* head)
The pointer is passed by value.
This sounds silly, it's a freaking pointer, right? Kind-of the definition of pass by reference. Well, the Node that's being pointed at is passed by reference. The pointer itself, head is just another variable that happens to contain the address of some other variable, and it's not being passed by reference.
So head contains a copy of the Node pointer used to call reverseLinkedList, and as with all parameters passed by value, any modifications to the copy, pointing head somewhere else, are not represented in the calling function.
The basic knowledge i have about pointers is that they point to a memory location and that they can be used for changing and extracting the values stored at that address.
for eg.
int a=5;
int *b;
b=&a;
Now here b points to the memory location of a.
cout<<*b; //gives the output 5
cout<<b; //outputs the address of a.
Basically for the purpose and passing the values by reference in functions and in data structures such as linked list and trees.
main()
{
node *head=NULL;// A pointer head of type node structure
add(&head,2); //adds the value to linked list
}
void add(node**head,int data)
{
//adds value into node
}
Can any one tell me why the &head has been received as **head in the add() function.
Also, what difference would it have made had it been received as *head?
when i print the value of **head it gives me a compiler error.
type of head is node*, when you get address of node* (&head), type of it is node**.
Difference between node** and node* is same as int* and int. One of them defines a pointer to other type.
add method gets a pointer because it might change the value of head. if *head is NULL, it should create a new node and assign head to address of this new node. if it inputted just node head, assigning head = malloc(...) wouldn't work since c is pass by value and it just assigns the "head" defined within the function, not head in the main. but if gets the address of the head in the main, it can freely alter it by accessing it like *head = malloc(...)
head is a pointer so you can't really print its value (well, you can if you cast it but it wouldn't be anything meaningful). You probably need to cast it
int *p; // p -> is a pointer to an int, *p -> integer contained in address pointed by p.
similarly
int **p2; // p2 -> a pointer to a pointer to an it, *p2 -> pointer to a int, **p2 int.
If you are confused, you can remember it as a covering rule. If you want to know what *p is, just cover *p ( we have int so it is an int ), if you to know what *p2 is, just cover it ( we have int * meaning it is a pointer to an int ).
This is my answer to another question: https://stackoverflow.com/a/20818342/2805305
The question was about an error the op got, but in my answer I explain exactly what you need to know.
The following codes are from the book Programming Interviews Exposed
I am having problem to understand the concept of pointers. Why can't we use the code no. 1.
Code no. 1
bool insertInFront( IntElement *head, int data )
{
IntElement *newElem = new IntElement;
if( !newElem ) return false;
newElem->data = data;
head = newElem; // Incorrect!
return true;
}
Code no. 2
bool insertInFront( IntElement **head, int data )
{
IntElement *newElem = new IntElement;
if( !newElem ) return false;
newElen->data = data;
*head = newElem; // Correctly updates head
return true;
}
In code 1, you are assigning the pointer location of newElem to the local copy of head. When the function returns, head is destroyed, and you will have a memory leak (you lose the pointer to newElem, leaving you with no way to delete it.) You've only changed the function's local copy of the pointer, the caller's copy is unaffected.
In code 2, head is a pointer-to-a-pointer. You don't have just a pointer, you actually have a pointer to the caller's pointer. This allows you to alter the caller's pointer, storing the pointer to newElem When the function returns, head is destroyed, but it's only the pointer-to-a-pointer. The original pointer is intact in the caller's scope.
Say you are calling function 1 as:
insertInFront(H, data);
Upon calling of a function, the computer makes duplication of arguments, then release them when the function returns. So in code No.1, head=newElem assigned the address of newElem to head(which is a duplicate of H), then released head, and the address of newElem is lost forever.
In code No.2, however, the function should be called as:
insertInFront(&H, data);
This means the ADDRESS of H is duplicated to head, and the address of newElem is assigned to *head, i.e. where head is pointing, which results in H. In this way you get the address of newElem after the function returns.
The concept of pointers are very complicated. Basically the way you need to think about pointers are that they are 'values' themselves. so:
*head = some_memory_address_that_points_to_head;
could be thought of as:
int im_a_pointer_to_head = some_memory_address_value_to_head;
So if we apply this same concept to your first function:
bool insertInFront(int im_a_pointer_to_head, int data)...
You are essentially passing in a copy of the value of your head pointer, changing it within the function only changes the temporary copy made for this function and does not actually change where the original pointer is pointing to.
The second function solves this because you are actually passing in the copy of the pointer that is pointing to the pointer that is pointing to the head (try saying that 3 times fast!). In this case, you are not changing the copied value, but instead are changing where the actual pointer is pointing to.
To change the value of anything passed via a function argument (and have that change persisted) outside of the function call, you MUST have the memory address of whatever it is whose value you want to change. This is another way of saying that you must "pass by reference" instead of "pass by value" -- otherwise pass by value will cause the function to just alter the COPY.
I want to NULL the head structure in case the head->next is NULL.
However it doesn't work when I pass it into a function to null it.
void remove(struct node* head)
{
int val;
cout << "Enter a value to delete: ";
cin >> val;
if (head->next == NULL)
if (head->data == val)
head = NULL;
}
Though it works fine when not passed to the function, but done directly in the main function.
Where do I go wrong?
The pointer passed as an argument to remove is being copied into the function. The head inside the function is a copy of that pointer. You set that copy to NULL and the one outside is left unchanged. You can simply take the pointer by reference:
void remove(struct node*& head)
{
// ...
}
Note the ampersand in the type of head. This means it's a "reference to pointer to node" and lets you reference precisely the object that is passed, rather than a copy of it.
The problem is that you are passing the pointer which lets you modify the node it points to, but you can't modify the pointer itself because the pointer itself is passed by value. So you can't do this in your remove function:
node = NULL;
A solution that people have given you is to pass it as a reference to the pointer, as in node *&head and this works perfectly fine. However I believe you'd like to know that the general convention for this sort of thing (modifying a pointer itself) is by passing a double pointer (i.e. a pointer to the pointer to node):
void remove(node **head) { ... }
Then in main you pass it the address of the pointer:
node *theHead = blah;
remove(&theHead);
Then in remove you can change the value of the pointer:
*head = NULL;
And you can also dereference it of course:
(*head)->next; // etc
If I pass a pointer P from function f1 to function f2, and modify the contents of P in f2, will these modifications be reflected in f1 automatically?
For example, if I need to delete the first node in a linked list:
void f2( Node *p)
{
Node *tmp = p;
p = p -> next;
delete tmp;
}
Will the changes made to P be reflected in the calling function, or will it now point to a memory space that has been deallocated?
( My intuitive answer here is no, changes are not reflected. However, good sources tell me that the above code will work. Can someone take the time to give the answer and explain the reasoning behind it also please? Also, if the above code is faulty, how can we achieve this without using a return type? )
If I pass a pointer P from function f1 to function f2, and modify the contents of P in f2, will these modifications be reflected in f1 automatically?
No. Since C and C++ implement pass-by-value, the value of the pointer is copied when it’s passed to the function. Only that local copy is modified, and the value is not copied back when the function is left (this would be copy-by-reference which a few languages support).
If you want the original value to be modified, you need to pass a reference to the pointer, i.e. change your signature of f2 to read:
void f2( Node*& p)
However, your f2 not only changes the pointer p, it also changes its underlying value (= the value being pointed to) by using delete. That change is indeed visible to the caller.
This will work, althrough not the way you want to.
If you call it like:
Node* mynode = new Node(); /*fill data*/
f2(mynode);
-> there will be the content of mynode undefined.
it will pass the value of the pointer to another function, and the value (p) will be local for function f2. If you want to modify it, use this way:
Node* mynode = new Node(); /*fill data*/
f2(&mynode);
void f2(Node** p)
{
Node* tmp = *p;
*p = (*p)->next;
delete tmp;
}