How to I switch this into using completely smart pointers? - c++

Question

I'm having trouble switching this linked stack file into using only smart pointers. I've tried some things and I just get loads of errors. Any help is appreciated. This is in C++ by the way.
There is supposed to be no raw pointers, no constructors not needed, and use of the auto keyword whenever possible.
Here's the code:
#include <new>
#include <string>
#include "PrecondViolatedExcep.h"
template <typename ItemType>
LinkedStack<ItemType>::LinkedStack(const LinkedStack<ItemType>& aStack) {
if (!aStack.topPtr) {
topPtr = nullptr;
}
else {
NodePtr origStackPtr(aStack.topPtr);
try {
topPtr = new Node<ItemType>(origStackPtr->getItem() );
NodePtr newStackPtr(topPtr);
origStackPtr = origStackPtr->getNext();
while (origStackPtr) {
newStackPtr->setNext(
new Node<ItemType>(origStackPtr->getItem())
);
newStackPtr = newStackPtr->getNext();
origStackPtr = origStackPtr->getNext();
}
}
catch (const std::bad_alloc&) {
while (!isEmpty() ) {
pop();
}
throw;
}
}
}
template <typename ItemType>
LinkedStack<ItemType>::~LinkedStack() {
while (!isEmpty() ) {
pop();
}
}
template <typename ItemType>
bool LinkedStack<ItemType>::isEmpty() const {
return !topPtr;
}
template <typename ItemType>
bool LinkedStack<ItemType>::push(const ItemType& newItem) {
try {
topPtr = new Node<ItemType>(newItem,
topPtr);
}
catch (const std::bad_alloc&) {
return false;
}
return true;
}
template <typename ItemType>
bool LinkedStack<ItemType>::pop() {
if(!isEmpty() ) {
NodePtr nodeToDeletePtr(topPtr);
topPtr = topPtr->getNext();
delete nodeToDeletePtr;
return true;
}
return false;
}
template <typename ItemType>
ItemType LinkedStack<ItemType>::peek() const {
// throw(PrecondViolatedExcep) {
if (isEmpty() ) {
std::string message("LinkedStack::peek() ");
message += "called on an empty stack.";
throw PrecondViolatedExcep(message);
}
return topPtr->getItem();
}

Answer 1

I hope the commentary on each piece of the answer demonstrates the principles well enough you can learn more than just these specific solutions.
Though you don't show the definitions of the class templates, it looks like LinkedStack<ItemType> has a private member topPtr, and you'd like to change its type from Node<ItemType>* to std::unique_ptr<Node<ItemType>>. Also, class template Node has member functions something like:
template <class ItemType>
class Node
{
// ...
// Either public, or Node befriends LinkedStack in some way.
explicit Node(const ItemType& value, Node* next = nullptr);
Node* getNext() const { return nextPtr; }
void setNext(Node* ptr) { nextPtr = ptr; }
// ...
// Probably private:
Node* nextPtr;
};
Changes to this Node template could look like:
template <class ItemType>
class Node
{
// ...
// Either public, or Node befriends LinkedStack in some way.
explicit Node(const ItemType& value)
Node(const ItemType& value, std::unique_ptr<Node> next)
: Node(value) { nextPtr = std::move(next); }
Node* getNext() const { return nextPtr.get(); }
void setNext(std::unique_ptr<Node> ptr) { nextPtr = std::move(ptr); }
void resetNext() { nextPtr = nullptr; }
[[nodiscard]] std::unique_ptr<Node> releaseNext()
{ return std::move(nextPtr); }
// ...
// Probably private:
std::unique_ptr<Node> nextPtr;
};
The member becomes a smart pointer because the Node "owns" the responsibility for cleaning up its "next" node, if any.
The pointer parameters of the two-argument constructor and setNext become smart pointers because calling each means the Node will take over responsibility for cleaning up that next Node. They need to use std::move to allow the member to actually take that responsibility from the parameter.
You might ask, shouldn't getNext return a smart pointer? No, because that would mean that calling getNext transfers responsibility for cleaning up the next node to the caller, and that's usually not what getNext is for. A raw pointer can still have its place, meaning a pointer to an object whose ownership is handled by something else, or possibly a null pointer. (When null isn't a possibility, we'd often consider changing from a raw pointer to a reference, which also implies ownership is handled by something else.)
Though for cases where we do want to take ownership back from a Node, I've added releaseNext(). It returns a unique_ptr to "give away" its responsibility, and in the process its own unique_ptr member becomes empty.
Finally, I've added resetNext as a way to reset the nextPtr back to null, more straightforward than passing an empty smart pointer to setNext. (And/or, we could overload a setNext(std::nullptr_t).)
Then Node can follow the Rule of Zero: It does not need to declare a destructor, copy constructor, move constructor, or any sort of assignment operator. Since it has a member of type std::unique_ptr<Node>, its implicitly-declared copy constructor and copy assignment members will be defined as deleted, meaning the compiler will complain about any code that tries to use them. But it will have working move constructor and move assignment operator, which are probably not needed but harmless.
Now to class template LinkedStack. It can't use the Rule of Zero since copying it should be allowed despite the unique_ptr member and should do a deep copy. So we'll go with the Rule of Five, as modified by the Copy and Swap Idiom:
template <class ItemType>
class LinkedStack
{
public:
LinkedStack() = default;
LinkedStack(const LinkedStack&);
LinkedStack(LinkedStack&&) = default;
LinkedStack& operator=(LinkedStack) noexcept;
~LinkedStack() = default;
friend void swap(LinkedStack& s1, LinkedStack& s2) {
using std::swap;
swap(s1.topPtr, s2.topPtr);
}
// ...
private:
std::unique_ptr<Node<ItemType>> topPtr;
};
The destructor is okay to default, so with the = default; above, you can delete your custom definition.
Assignment is defined per Copy And Swap:
template <class ItemType>
LinkedStack<ItemType>& LinkedStack<ItemType>::operator=(
LinkedStack rhs) noexcept
{
swap(*this, rhs);
return *this;
}
The copy constructor gets simpler:
template <typename ItemType>
LinkedStack<ItemType>::LinkedStack(const LinkedStack<ItemType>& aStack)
: topPtr() // initially null
{
if (aStack.topPtr) {
Node* origStackPtr = aStack.get();
topPtr = std::make_unique<ItemType>(origStackPtr->getItem());
Node* newStackPtr(topPtr.get());
origStackPtr = origStackPtr->getNext();
while (origStackPtr) {
newStackPtr->setNext(
std::make_unique<ItemType>(origStackPtr->getItem())
);
newStackPtr = newStackPtr->getNext();
origStackPtr = origStackPtr->getNext();
}
}
}
The loop uses raw Node* pointers because once the nodes are safely stored in the topPtr or in another Node, the constructor code no longer needs to worry about deleting them. But it does need to reassign the variables as the loop executes, so references won't do, and it also needs to detect when origStackPtr->getNext() returns a null pointer.
Your original needed the try-catch-rethrow because in case of an exception in a constructor body, the destructor for that class is not called. But destructors of its members and base classes are called. So now if an exception happens in the copy constructor body, the destructor for the unique_ptr member topPtr executes, which will take care of deleting the top Node. Destruction of that Node's unique_ptr member will likewise delete its next node if any, and so on recursively - all with zero lines of (your) code.
isEmpty does not need any change: the expression !topPtr also works with a unique_ptr, meaning "is not null".
A straightforward update of push just changes the new to a make_unique:
template <typename ItemType>
bool LinkedStack<ItemType>::push(const ItemType& newItem) {
try {
topPtr = std::make_unique<Node<ItemType>>(
newItem, topPtr);
}
catch (const std::bad_alloc&) {
return false;
}
return true;
}
However, I'd recommend getting rid of the try/catch and return value. A memory error is rare, so most code shouldn't be checking to see whether this push called one. Something that really does care ought to do a try/catch itself around whatever amount of code is most appropriate. This would reduce the body to just one statement.
And pop gets simpler:
template <typename ItemType>
bool LinkedStack<ItemType>::pop() {
if (!isEmpty()) {
topPtr = topPtr->releaseNext();
return true;
}
return false;
}
Notice in the topPtr reassignment statement, first releaseNext() is used to take responsibility for the second Node (if any) away from the top Node. Then that responsibility is immediately given to the topPtr variable by the assignment. This also means topPtr will delete what it previously pointed at, the top node being popped off. Since that top node no longer has a next node, nothing else gets deleted, and the stack ends up in the correct state just like in the old version.
peek() does not require any change. The topPtr->getItem() expression will work via the unique_ptr<T>::operator-> function. (But I'd suggest changing the return types of Node<ItemType>::getItem() and LinkedStack<ItemType>::peek() to const ItemType&, to avoid useless copies when ItemType is a class type.)