The head node in a linked list - c++

Question

The following naive code implements a linked list, without printing all the elements in the main function, everything would be fine. However, the LinkedList::printll function will trigger a set fault(Gcc 5.3.0), the problem is related to the appropriate handling of the head node I suppose...
So, is there any way to make this code work with least modification of the printll function?
#include <iostream>
using namespace std;
struct Node{
int value;
Node* next;
};
struct LinkedList{
Node* head= NULL ;
void append(int);
void printll();
};
void LinkedList::append(int data){
Node* cur = head;
Node* tmp = new Node;
tmp->value = data;
tmp->next = NULL;
if(!cur){
cur = tmp; // cur-> head
}
else{
while(cur->next != NULL){
cur = cur->next;
}
cur->next = tmp;
}
std::cout<<cur->value<<std::endl; // cur-> temp
delete tmp; // comment out
}
void LinkedList::printll(){
Node* cur = head;
while(cur->next != NULL){ //
std::cout<<cur->value<<std::endl;
cur = cur->next;
}
}
int main(){
LinkedList LL;
LL.append(5);
LL.append(6);
LL.append(7);
LL.printll(); // --without this, the program is fine
return 0;
}

Answer 1

You have some bugs in append:
if(!cur){
cur = tmp;
}
This only assigns to the local copy. I assume you are trying to set head here, so do that: head = tmp;. Note that in this case, you can't print cur, since you haven't set it. You could print tmp->value though.
Then:
delete tmp;
You only just created it and assigned it into place - why are you deleting it? You know that there is still a pointer to it. Only delete it when you come to clean up the list when you are done with it (which you don't do at all at the moment).
Other than that, your printll won't print the last element - think about when it will stop:
A -> B -> C -> NULL
It will stop on node C, but never print C's value. You can just replace:
while(cur->next != NULL){
with
while(cur != nullptr){
(Also, I don't like endl).
See here for these changes running:
#include <iostream>
struct Node{
int value;
Node* next;
};
struct LinkedList{
Node* head = nullptr ;
void append(int);
void printll();
};
void LinkedList::append(int data){
Node* cur = head;
Node* tmp = new Node;
tmp->value = data;
tmp->next = nullptr;
if(!cur){
head = tmp;
}
else{
while(cur->next != nullptr){
cur = cur->next;
}
cur->next = tmp;
}
}
void LinkedList::printll(){
Node* cur = head;
while(cur != nullptr){
std::cout << cur->value << '\n';
cur = cur->next;
}
}
int main(){
LinkedList LL;
LL.append(5);
LL.append(6);
LL.append(7);
LL.printll();
}

Answer 2

1.you cann't
delete tmp;
cause tmp is a pointer, when you run delete tmp, you delete the object.
2.the print function should like this:
void LinkedList::printll(){
Node* cur = head;
while(cur->next != NULL){ // -> problems is here
std::cout<<cur->value<<std::endl;
cur = cur->next;
}
std::cout<<cur->value<<std::endl;
}