Re: [Boost-users] shared_ptr and typedef

25 Jan 2006


      ...
...
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 
#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

Re: [Boost-users] shared_ptr and typedef

Sohail Somani