如何使用typdef模板化的结构指针？

Question

Here is my code:

#include<iostream>
using namespace std;

template<typename T>
struct Doublenode{
    T data;
    struct Doublenode<T>* prev;
    struct Doublenode<T>* next;
};
template<typename T>
using Double_node = typename Doublenode <T>::Double_node;
template<typename T>
using DoubleNodePtr = typename Double_node <T>* ::DoubleNodePtr;

This gives me the folowing error:

error: expected nested-name-specifier using DoubleNodePtr = typename (^)Double_node * ::DoubleNodePtr;

What is the right way to typdef a templated struct pointer?

Answer 1

The syntax of a using type alias, be it templated or not, is as follows:

using <alias name> = <existing type name>;

You are writing it as

using <alias name> = <existing type name> :: <alias name>;

It could work if <existing type name> has a member type named like <alias name> . But it's not correct in general. You problem, ultimately, seems to be about misunderstanding the required syntax. Which when applied to your post is simply:

template<typename T>
using Double_node = Doublenode<T>;

Answer 2

 template<typename T> using DoubleNodePtr = typename Double_node <T>* ::DoubleNodePtr; 

Pointer types do not have members or a scope, so you cannot apply the scope resolution operator on a pointer.

 template<typename T> using Double_node = typename Doublenode <T>::Double_node; 

Doublenode doesn't have a member called Double_node , so this can only be sensible if you intend to specialize Doublenode and those specializations have that member.

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C++ - How to typdef templated struct pointer?

Like this:

template<typename T>
using DoubleNodePtr = Double_node<T>*;

However, it is usually a bad idea to create a type alias for a pointer. Don't do it.

Answer 3

template<typename T>
using Double_node = Doublenode<T>;

template<typename T>
using DoubleNodePtr = Double_node<T>*;