template default

scaloni (5)
Hello,
i wanna know if is possible declare a class(A for ex) and use it with template default on your own.

1
2
 
template <class T, class U = A>
class A { .... }


If it`s possible, how can I do that ?
thank`s
Zhuge (4664)
I don't see why not, so long as the A has been forward declared. Obviously, you couldn't make an actual instance of A inside of A, but you could make a pointer or something like that.
scaloni (5)
I got errors doing this.
Can you do an example and post here?
LB (13399)
You'd have to give it the first template parameter like this:
1
2
 
template<typename T, typename U = A<int> >
class A { /* * */ };
scaloni (5)
If I try this:
1
2
3
4
5
6
7
8
9
10
 
template<typename T, typename U = A<T> >
class A { 
	A<T> *ex;
};

int main(void){
	
	A<int> a;
	
};


and I get this:

1
2
3
4
5
6
7
8
 
teste.cpp:8:35: error: expected type-specifier before 'A'
teste.cpp:8:35: error: expected '>' before 'A'
teste.cpp:10:5: error: template argument 2 is invalid
teste.cpp: In function 'int main()':
teste.cpp:15:7: error: template argument 2 is invalid
teste.cpp:15:10: error: invalid type in declaration before ';' token
teste.cpp:15:10: warning: unused variable 'a'
Zhuge (4664)
Did you try forward declaring your class? A might not yet exist inside the template<> specifications for it.
scaloni (5)
What is the best way for forward declaring?

I did this:
1
2
 
template<typename T, typename U>
class A;

But I don't know how to use a prototype in this case.
Thanks.
Last edited on
Zhuge (4664)
That looks like a fine forward declaration to me.
aquaz (170)
What are you trying to do exactly?
Forward declaration or not, the problem here is the infinite template recursion.
You cant stop it by specifying a second template argument like this:
1
2
3
4
5
6
7
8
9
10
 
template<typename T, typename U = A<T, int> >
class A { 
	A<T> *ex;
};

int main(void){
	
	A<int> a;
	
};
closed account (zb0S216C)
+ 1 to Aquaz.

Wazzak
scaloni (5)
So, not is possible I do something like this: 
1
2
3
4
5
6
7
8
9
10
 
template<typename T, typename U = A<T, A> >
class A { 
	A<T> *ex;
};

int main(void){
	
	A<int> a;
	
};

Because the template U must to be A and not int
Thank you
Topic archived. No new replies allowed.