aspen100 aspen100 - 2 months ago 14
C++ Question

What is the meaning of template arguments of this type - `ClassA<T> T::*ELEM`?

I came across this code recently, in the context of intrusive lists:

template<typename T> struct Node{
T *next;
T *prev;
Node(): next(nullptr), prev(nullptr){}

* Intrusive doubly-linked-list
* */
template<typename T, Node<T> T::*NODE>
class List{
T *head;
T *tail;
List():head(nullptr), tail(nullptr){}
~List() {clear();}

* Add an element at the head of list
* @param elem item to be inserted
* */
void add_to_front(T *elem){

Node<T> *node = &(elem->*NODE);

assert((node->next) == nullptr);
assert((node->prev) == nullptr);

node->next = head;

if(head != nullptr){
Node<T> *temp = &(head->*NODE);
temp->prev = elem;

head = elem;

if(tail == nullptr)
tail = head;

//other member functions ,etc.

I see similar code in the boost intrusive list library (
member_hook<class T, class Hook, Hook T::* PtrToMember>

My question is regarding the template argument
Node<T> T::*NODE
. I am not an expert in C++, but I have never come across this particular syntax before, and don't know what to search for to understand it.

What does it mean? What is its purpose, and what should I interpret it as - "NODE is a pointer to a Node, belonging to T"? That doesn't make sense to me, because T isn't known to contain specific members ahead of time, and as far as I know,
is used to resolve scope.

As well, if someone could clarify the use of
in this line, for instance :
Node<T> *node = &(elem->*NODE);
, that would help me understand what this is being used for.


It's a member pointer. NODE is a pointer to a member of T that has type Node<T>.

Similarly, foo->*bar is shorthand for (*foo).*bar where .* is the dereference-pointer-to-member operator. elem->*NODE accesses the member of *elem pointed to by NODE.