shared_ptr with templates

Question

Hello!

If I want to create a smart pointer to struct I do that:

 struct A
 {
  int value;
 };
 typedef boost::shared_ptr<A> A_Ptr;

So, I can write the following:

 A_Ptr pA0(new A);
 pA0->value = 123;

But, If I have a template struct like that:

 template<typename T>
 struct B
 {
  T value;
 };

And I want to write the following:

 B_Ptr<char> pB0(new B<char>);
 pB0->value = 'w';

So, How should I declare the B_Ptr ?

Answer 1

If you are interested in a fixed template type for B, then I throw my support behind xtofl's answer. If you're interested in later specifying B's template argument, C++ doesn't allow you to do this (though it will be changed in C++0x). Typically what you're looking for is this kind of workaround:

template <typename T>
struct B_Ptr
{
    typedef boost::shared_ptr< B<T> > type;
};

B_Ptr<char>::type pB0 = ...;

(Thanks to UncleBens for the improvements.)

Answer 2

That would be

typedef shared_ptr< B<char> > B_Ptr;
B_Ptr p( new B<char> );
p->value = 'w';

Answer 3

What you want is not yet possible in C++. You want "template typedefs" which will be known in C++0x as "alias declaration templates":

template<typename T>
struct A {};

template<typename T>
using APtr = boost::shared_ptr<A<T>>;  // <-- C++0x

int main() {
    APtr<int> foo;
}

I guess you could do something similar in C++98 with a macro if you really want to.