Re: [Boost-users] shared_ptr and typedef

From: Sohail Somani (s.somani_at_[hidden])
Date: 2006-01-25 14:14:24

> > Why is there a need? Presumably you're potential surrogate 
> for shared_ptr  
> > will also support operator *, and that's all you use in 
> your function.
> 
> Why is there a need to do what?  I'm not following you here.  I've
> gone through and specified Type everywhere in my application.  It's
> ugly but it works.  Would the upcoming template typedef feature
> avoid the need to specify Type in the call?

I think what Jens is getting at is that you should use the duck-typing
that 
C++ templating affords you:

#include <boost/smart_ptr.hpp>
#include <iostream>

using namespace std;

template<typename Handle>
void print_handle(Handle p)
{
        cout << *p << endl;
}

int main()
{
        int * a = new int(5);
        boost::shared_ptr<int> b(new int(6));

        print_handle(a);print_handle(b);
}

$ ./a.out
5
6