I have written a class with protected constructor, so that new instances can only be produced with a static create() function which returns shared_ptr's to my class. To provide efficient allocation I'd like to use boost::make_shared inside the create function, however the compiler complains that my class constructor is protected inside boost::make_shared. I decided to my boost::make_shared a friend of my class but I'm puzzled about the syntax. I tried
template< class T, class A1, class A2 >
friend boost::shared_ptr<Connection> boost::make_shared(const ConnectionManagerPtr&, const std::string&);
but the compiler gived me syntax errors. Please help.
You don't need to template the friend
part, but you need to signify that the friend
function is a template:
friend boost::shared_ptr<Connection> boost::make_shared<>(/* ... */);
// ^^
That works with Comeau and current GCC versions but fails with VC. Better would be the following form:
friend boost::shared_ptr<Connection> boost::make_shared<Connection>(/* ... */);
That works across multiple compilers now - i tested it on VC8, VC10, GCC 4.2, GCC 4.5 and Comeau 4.3.
Alternatively using a qualified name to refer to a particular instance of the function template as Martin does should work and does with Comeau, but GCC chokes on it.
A useful alternative that doesn't depend on the implementation details of make_shared()
(and thus also works with VC10s TR1 implementation) is to use the pass-key-idiom for access-protection of the constructor and to befriend the create()
function instead, e.g.:
class Connection {
// ...
public:
class Key {
friend boost::shared_ptr<Connection> create(const ConnectionManagerPtr&,
const std::string&);
Key() {}
};
Connection(const ConnectionManagerPtr&, const std::string&, const Key&);
};
boost::shared_ptr<Connection> create(const ConnectionManagerPtr& p,
const std::string& s)
{
return boost::make_shared<Connection>(p, s, Connection::Key());
}