Reputation: 602
everyone! I am making my own linked list template to both practice and for future use; however, I ran into a problem with one of my functions:
Node* LinkedList::FindNode(int x); //is meant to traverse the list and return a pointer to the containing x as its data.
When trying to declare it in my implementation file, I keep getting messages of Node being undefined and incompatibility errors.
Here is my header file:
#pragma once
using namespace std;
class LinkedList
{
private:
struct Node
{
int data;
Node* next = NULL;
Node* prev = NULL;
};
//may need to typedef struct Node Node; in some compilers
Node* head; //points to first node
Node* tail; //points to last node
int nodeCount; //counts how many nodes in the list
public:
LinkedList(); //constructor
~LinkedList(); //destructor
void AddToFront(int x); //adds node to the beginning of list
void AddToEnd(int x); //adds node to the end of the list
void AddSorted(int x); //adds node in a sorted order specified by user
void RemoveFromFront(); //remove node from front of list; removes head
void RemoveFromEnd(); //remove node from end of list; removes tail
void RemoveSorted(int x); //searches for a node with data == x and removes it from list
bool IsInList(int x); //returns true if node with (data == x) exists in list
Node* FindNode(int x); //returns pointer to node with (data == x) if it exists in list
void PrintNodes(); //traverses through all nodes and prints their data
};
If someone can help me define a function that returns a Node pointer, I would greatly appreciate it!
Thank you!
Upvotes: 0
Views: 13295
Reputation: 65126
Since Node
is declared within another class, did you remember to include the class name when referring to it in your implementation?
LinkedList::Node *LinkedList::FindNode(int x) { ... }
In the class declaration the prefix isn't required because the declaration is inside the class, and therefore Node
is implicitly available.
Upvotes: 3