How to use typedef together with template ?

[VanKha]

I have some class,e.g

template<class T>
class node{
  T key;
  node* next;
};

and I want to define a pointer to the class node as a type . How to declare ?
I try typedef node<T>* nodeptr but it doesn't work,there'll be error .

[JosAH]

I have some class,e.g

template<class T>
class node{
  T key;
  node* next;
};

and I want to define a pointer to the class node as a type . How to declare ?
I try typedef node<T>* nodeptr but it doesn't work,there'll be error .

What is the error exactly? That error surprises me a bit ...

kind regards,

Jos

[weaknessforcats]

A typedef creates a new name for an existing type. A class template is not a type. Rather, it is a pattern for a type. Not until you specialize it can you typedef it:
[code=cpp]
typedef node<T>* nodeptr; //error
typedef node<int>* nodeptr; //OK
[/code]

[JosAH]

A typedef creates a new name for an existing type. A class template is not a type. Rather, it is a pattern for a type. Not until you specialize it can you typedef it:
[code=cpp]
typedef node<T>* nodeptr; //error
typedef node<int>* nodeptr; //OK
[/code]

Darn, I missed this one completely; thanks for helping out.

kind regards,

Jos

[VanKha]

Thanks a lot . But if I have to declare a specific T , another problem arises . It is because of the fact that I just compiled successfully a program with the class node and a class list using that "node" . I do like this

class node...
typedef node* nodeptr;
class list
//..using nodeptr many times here

and it compiles well !
Then I decide to use the template . And so , I must have node<T>* as nodeptr to make use of it in the list class . If I cannot declare nodeptr like the wrong syntax(!!!) , how can I write the list class ?Don't tell me that just go around the program and replace nodeptr with node<T>* ! Is there anyway to come around this problem ?

[weaknessforcats]

Then I decide to use the template . And so , I must have node<T>* as nodeptr to make use of it in the list class .

You mean like this:
[code=cpp]
template<class T>
class node{
T key;
node* next;
};


template<class T>
struct List
{
node<T>* start;
node<T>* end;
node<T>* current;
};

int main()
{
List<int> mylist;
}
[/code]

This compile and links just fine.

No typedef.

[VanKha]

Yes,but the thing is I have written nodeptr in my program instead of node<T>* ! I mean that at first I wrote like that and after updating with the template method pattern , I had to change all to node<T>* ? If the typedef could have worked , I have to do nothing but now that the typedef is wrong,I have to change all !! Sorry,just because I think that it's so unreasonable . Have anybody encounter the problem like me ? I think the answer must be yes,because just like me now,everybody must have the 'newbie in c++' period , so after writing the code without template,then reading textbooks,seein g the usefulness,then attemping to apply the template method,you'll expect that just change from typedef node* nodeptr to typdef node<T>* nodeptr and no more ! It's really natural expectation,isn 't it ?

[weaknessforcats]

It's expected with a typedef that you already have a type. A template is not a type.

Ultimately, you will need to change your code so it is acceptable to the compiler. You aren't the first one who had to ditch a lot of work over an assumption.

The lesson here is to write only a little code and then compile. Once it compiles, then write a little more. That way you always have code that compiles and any errors will usually be in the code that was last added.

Writing code for a long time and then doing a compile is a recipe for rework and frustration.