Reputation: 589
I originally had to create my own linked list with using the STL. Now, I am to implement a copy constructor method and I am having real difficulty understanding it. Have a test on this in a few days so I would really like to get it clear. (Test is closed book so need to really). The List contains a EmployeeNode pointer *head. The EmployeeNode contains an Employee and a pointer to the next EmployeeNode. The Employee class contains a name and salary.
The method seems to get caught in the for loop when trying to copy the 3rd node over. I think this is because I overwrite the newNode but I do not know how to solve this.
ListOfEmployee::ListOfEmployee(const ListOfEmployee &obj)
{
head = NULL;
if(obj.head != NULL)
{
EmployeeNode *newNode = new EmployeeNode("", 0);
EmployeeNode *tempPtr;
EmployeeNode *newPtr;
//using the temp pointer to scroll through the list until it reaches the end
for(tempPtr = obj.head; tempPtr->next !=NULL; tempPtr = tempPtr->next)
{
if(head == NULL)
{
cout<<"Attempts to initialize the head"<<endl;
head = newNode; //assinging the new node to the head
newNode->emp.name = tempPtr->emp.name;
newNode->emp.salary = tempPtr->emp.salary;
cout<<"Initializes the head"<<endl;
}
else
{
cout<<"Attempts to add a new node"<<endl;
//using the temp pointer to scroll through the list until it reaches the end
for(newPtr = head; newPtr->next !=NULL; newPtr = newPtr->next)
{
cout<<"Looping through the list"<<endl;
}
//assiging the last place to the new node
newPtr->next = newNode;
newNode->emp.name = tempPtr->emp.name;
newNode->emp.salary = tempPtr->emp.salary;
cout<<"Adds a new node"<<endl;
}
}
}
}
Upvotes: 0
Views: 4141
Reputation: 704
In your code where you are adding newNode in newPtr->next = newNode;
you are basically using previously allocated node. You should create a new node using new. Something like:
newPtr->next = new EmployeeNode("", 0);
newNode = newPtr->next;
newNode->emp.name = tempPtr->emp.name;
newNode->emp.salary = tempPtr->emp.salary;
Also you should set newNode->next = NULL;
in your code.
Upvotes: 1