Thank you. I understand that.

ps: It's my mis-express that  the implicit conversion from T* to shared_ptr<T>.
 I mean that  Boost.Python constructs a shared_ptr<T> by it's  construction method with a raw pointer .

2008/8/5 Peter Dimov <pdimov@pdimov.com>
Not exactly, but the end result is the same. Python does pass a raw pointer (PyObject*) to the Boost.Python translation layer. Boost.Python then constructs a specially crafted shared_ptr<A> from that. There is no implicit conversion from T* to shared_ptr<T>.
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