C++ define alias for a struct using template - c++

Question

using c++ language : i made a structure generic and i want to give it an alias but failed !
here is the code:
template<class T>
struct noeud {
T val;
struct noeud *next;
};
typedef struct noeud * Noeud;
here is the error:
template argument required for 'struct noeud'
invalid type in declaration before ';' token
Build : 2 error(s), 0 warning(s)

Answer 1

you can achieve this by using a syntax introduced in the C++11 revision :
template<class T>
struct Node;
template <class T>
using NodePtr = Node<T>*;
You should stay away from naked pointer and use a smart pointer instead, like unique_ptr :
template <class T>
using NodePtr = std::unique_ptr<Node<T>>;

Answer 2

it should be
typedef struct noeud<your type>* Noeud;
where your type is the type you want the struct to be instanciated with

Answer 3

If you want to have another template as an alias you have to use the using keyword:
template<typename T>
using AliasName = noeud<T>*;

Answer 4

To add to the other answers:
The class noeud is a template and thus it needs a template parameter to become a type as template types are deduced at compile time. E.g.:
template<class T>
struct noeud {
T val;
//noeud* next // Error: compiler doesn't know what a noeud is.
noeud<T>* next; // Need to specify template parameter (T can be used).
};
typedef noeud<int>* Noeud; // Need to specify template parameter.
Also in C++ you don't have to put the keyword struct in front of the type when declaring "structs"/aggregate classes when there is no ambiguity. In C++ the keyword struct means class with default access specifier set to public.
As mentioned by others, in C++11 you can use the using syntax to declare a templated alias (typedef does not allow templated aliases), e.g.:
template<typename T>
using noeud<T>* Noeud; // Use 'using' to create an alias in same manner as typedef.
This allows you to use the alias as:
Noeud<int> n; // Type is: noeud<int>*