C++ -- Is there an implicit cast here from Fred* to auto_ptr<Fred>?

Question

I saw the following code,

#include <new>
#include <memory>
using namespace std;

class Fred;  // Forward declaration
typedef  auto_ptr<Fred>  FredPtr;

class Fred {
public:
  static FredPtr create(int i)
  { 
    return new Fred(i); // Is there an implicit casting here? If not, how can we return
                        // a Fred* with return value as FredPtr?
  }
private:
  Fred(int i=10)      : i_(i)    { }
  Fred(const Fred& x) : i_(x.i_) { }
  int i_;
};

Please see the question listed in function create.

Thank you

// Updated based on comments

Yes, the code cannot pass the VC8.0 error C2664: 'std::auto_ptr<_Ty>::auto_ptr(std::auto_ptr<_Ty> &) throw()' : cannot convert parameter 1 from 'Fred *' to 'std::auto_ptr<_Ty> &'

The code was copied from the C++ FAQ 12.15.

However, after making the following changes,

replace 
  return new Fred(i);
with
  return auto_ptr<Fred>(new Fred(i));

This code can pass the VC8.0 compiler. But I am not sure whether or not this is a correct fix.

Answer 1

std::auto_ptr does have a constructor that takes a raw pointer as its argument, but that constructor is explicit and cannot be used as a converting constructor.

This code will not compile.