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.
Related
I'm just a beginner at C++ and today while we are learning about linked list my teacher showed us how to delete from front at the linked list.The problem is I didn't understand why we are deleting pointer p which is static memory can't we just do it with second code that uses dynamic memory pointer which is temp? Head is dynamic memory pointer for object of type Node
//My teacher code
template <class T>
void DeleteFront(Node<T>* & head)
{
// save the address of node to be deleted
Node<T> *p = head;
// make sure list is not empty
if (head != NULL)
{
// move head to second node and delete original
head = head->NextNode();
delete p;//I didn't understand this line because our p declaration is static
}
}
//My suggestion
template <class T>
void DeleteFront(Node<T>* & head)
{
// save the address of node to temporary dynamic pointer
Node<T> *temp ;
temp=new Node<T>(head);
// make sure list is not empty
if (head != NULL)
{
// move temp to second node which will be showed by head
temp = head->NextNode();
delete head;//delete front item
head=temp;//assign the address of second node to head
}
}
p is not declared static. In fact it's a local variable. Static variables can only be declared in class scope, and are marked with the keyword static.
A static variable is a variable that is accessible independently of any objects and is valid for the class itself, having the same value for any object, or even without any objects. This is not the case, it's a very normal local variable, which resides on the stack and holds the pointer to a Node object.
When you call delete on that pointer, you delete what the pointer points to, not the pointer object on the stack itself. The pointer itself of course also has an address, which you would get with &p and could be saved into a Node<T>** variable. Though there is no need to do that. Calling delete on an object that is on the stack doesn't work.
Now regarding the code:
You are creating a new node with
Node<T> *temp ;
temp=new Node<T>(head);
which is not only unnecessary but in fact you are leaking memory, since you override the pointer value or exit the function without calling delete.
Every new should have a corresponding delete, at least when the pointer goes out of scope, such as the end of your function. The line is completely superfluous. A better way to initialize your a pointer is:
Node<T> *temp = nullptr;
Other than that, your code does the same thing. your teacher saves what is to be deleted into p, which works but might seem unintuitive, while you save what is to be saved (head->NextNode()). Both work.
Also NextNode() is a function and should have a lower case name. A function's name should also imply what it does. While this is easy enough to know in this case, NextNode isn't really a verb/action. getNextNode() would be a better name.
To improve on your teacher's code you could put the declaration of p into the if-block and save the p = head as well as the stack operations if the list is not empty, but I'm sure the compiler would do that for you.
Node<T> *p = head;
// make sure list is not empty
if (head != NULL)
{
// move head to second node and delete original
head = head->NextNode();
delete p;
}
This section of code assigns the pointer p to the head passed into it. It then checks if the head is NULL, and if it is not then it moves the head to the next node before deleting p -- which is not a static. It is local, but not static. This removes the original head from memory and removes most chances at a dangling pointer.
Node<T> *temp ;
temp=new Node<T>(head);
if (head != NULL)
{
temp = head->NextNode();
delete head;
head=temp;
}
This code deletes the original head after assigning the head to a temp object, then passes the object back to head. It does not delete the temp Node object. This would be a potential dangling reference, as the head is getting passed by address and the original address was never deleted when held by temp.
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
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.
I have a school assignment which requires me to create a binary tree in c++, complete with all the usual overloaded operators (=, ==, !=, copy, destroy, etc). I am currently struggling to program the copy constructor method. In this assignment, I am being asked NOT to use the binary tree class' insert method in either the operator= or copy constructor methods. Here's the binary tree class header file:
#ifndef BINTREE_H
#define BINTREE_H
#include <iostream>
#include "nodedata.h"
using namespace std;
class BinTree
{
public:
BinTree(); // constructor
BinTree(const BinTree &); // copy constructor, calls copyHelper
~BinTree(); // destructor, calls makeEmpty
bool insert(NodeData*); // insert method, inserts new nodes
bool isEmpty() const; // true if tree is empty, otherwise false
private:
struct Node
{
NodeData* data; // pointer to data object
Node* left; // left subtree pointer
Node* right; // right subtree pointer
};
Node* root; // root of the tree
Node& copyHelper(const Node*); // copies the tree recursively
void makeEmpty(Node*); // deletes nodes recursively
};
#endif
And here's what I've got so far for the copy constructor:
BinTree::BinTree(const BinTree &otherTree)
{
root = copyHelper(otherTree.root); // 1
}
Node& BinTree::copyHelper(const Node* other) // 2
{
if(other == NULL)
{
return NULL; // 3
}
Node newNode; // 4
if(other.data == NULL)
{
newNode.data = NULL; // 5
}
else
{
NodeData newNodeData(*other->data); // 6
newNode.data = newNodeData; // 7
}
newNode.left = copyHelper(other.left); // 8
newNode.right = copyHelper(other.right); // 9
return newNode; // 10
}
This is causing all manner of compile errors, and I don't understand any of them. Here's what I THINK should be happening:
•Overview: The entire tree will be copied from the ground up recursively. Each node should already contain data and links to all subsidiary nodes (if they exist) when it is returned to the node above it.
Since root is, by definition, a pointer to a Node object, there should be no issue with assigning it to a function which returns a Node object. However, the compiler is giving me a conversion error here.
The copyHelper function, which is called recursively, takes a pointer to one of the original nodes as an argument, and returns a copy of this node.
If there is no original node, then there's no point in building a copy of it, so a NULL value is returned.
newNode will eventually become a copy of "other".
If "other" does not have any NodeData linked to it, then newNode's data pointer will link to NULL instead.
Otherwise, a new NodeData called "newNodeData" will be created, using the NodeData copy constructor, which is called by dereferencing other's NodeData pointer.
newNode's data pointer now points to newNodeData, which now contains the same string as other's NodeData object.
newNode's left pointer will point to another node, which is created recursively by calling copyHelper on whatever Node other's left pointer is assigned to.
newNode's right pointer will point to another node, which is created recursively by calling copyHelper on whatever Node other's right pointer is assigned to.
Only once this node and all the nodes beneath it have been copied and returned to it will it be returned itself.
Here are my existing questions:
I'm assuming that we only need to use -> instead of . when we're dereferencing a pointer. Is this correct?
I know sometimes we need to use the "new" statement when creating objects (see part 4), but I've typically only seen this done when creating pointers to objects. When, specifically are we supposed to be using the "new" statement?
If I were to create newNode as a pointer (IE: Node* newNode = new Node;), wouldn't this cause a pointer to a pointer when it was returned? Wouldn't that be less efficient than simply having the first pointer point directly to the returned Node object? Is there some technical reason I can't do it this way?
When is it advisable to take a pointer by reference as a parameter, instead of simply taking the pointer, or even the object itself? Is this relevant here?
The compiler seems to think I'm declaring a Node named BinTree::copyHelper instead of defining a function that returns a Node. How can I prevent this?
In general, I think I have the concepts down, but the syntax is completely killing me. I literally spent all day yesterday on this, and I'm ready to admit defeat and ask for help. What mistakes do you guys see in my code here, and how can I fix them?
This line:
Node newNode; // 4
Allocates a Node on the stack. When the function returns, the Node is no longer valid. It has gone out of scope. It may work for a while if another function call doesn't rewrite the stack.
You need to do a New on the node.
Node& copyHelper(const Node*); // copies the tree recursively
Must that be a reference? It looks like it should be Node* instead.
In your declaration of copyHelper, the type Node is nested within BinTree but you fail to implement this. It should instead be
BinTree::Node* BinTree::copyHelper(const BinTree::Node* other)
Yes, -> dereferences a pointer. It's a shortcut for *pointer.member.
new creates object on the heap and returns a pointer - it does not create a pointer. Every time you create a object on the heap, you must use new.
Node * newNode = new Node; assigns pointer to newly created Node to newNode. Node * is pointer to object, not pointer to pointer, which would be Node **. You cannot do Node newNode = new Node;, as Node * (pointer) is not convertible to Node (object).
When you take parameter as reference, you are (generally) guaranteed that the parameter is not not. You also does not need to dereference the reference and you can just use . instead of ->.
"The compiler seems to think I'm declaring a Node named BinTree::copyHelper instead of defining a function that returns a Node. How can I prevent this?" - how did you come to such a conclusion?
After much fretting and pulling of hair, I have rewritten my copy constructor and gotten it to compile. There may well still be errors in it, but I'm feeling a lot better about it now thanks in part to the tips posted here. Thanks again! Here's the modified code:
BinTree::BinTree(const BinTree &otherTree)
{
root = copyHelper(otherTree.root); // Starts recursively copying Nodes from
// otherTree, starting with root Node.
}
BinTree::Node* BinTree::copyHelper(const Node* other) // Needed BinTree at beginning
// due to nested Node struct
// in BinTree class.
{
if(other == NULL)
{
return NULL; // If there's no Node to copy, return NULL.
}
Node* newNode = new Node; // Dynamically allocated memory will remain after
// function is no longer in scope. Previous newNode
// object here was destroyed upon function return.
if(other->data == NULL) // Other is a pointer to data, which is also a pointer.
// -> dereferences other, and provides the item at the
// memory address which other normally points to. It
// just so happens that since data is a pointer itself,
// it can still be treated as such. I had stupidly been
// attempting to use . instead of ->, because I was afraid
// the -> would dereference data as well. If I actually
// wanted to DO this, I'd have to use *other->data, which
// would dereference all of other->data, as if there were
// parenthesis around it like this: *(other->data).
// Misunderstanding this was the source of most of the
// problems with my previous implementation.
{
newNode->data = NULL; // The other Node doesn't contain data,
// so neither will this one.
}
else
{
NodeData* newNodeData = new NodeData; // This needed to be dynamically
// allocated as well.
*newNodeData = *other->data; // Copies data from other node.
newNode->data = newNodeData;
}
newNode->left = copyHelper(other->left); // Recursive call to left node.
newNode->right = copyHelper(other->right); // Recursive call to right node.
return newNode; // Returns after child nodes have been linked to newNode.
}