eliminate one form of PythonObject::Reset()

[lawrence_danna]

Summary

I'd like to eliminate all forms of Reset() and all public constructors
on these objects, so the only way to make them is with Take<> and Retain<>
and the only way to copy or move them is with actual c++ copy, move, or
assignment.

This is a simple place to start.

[lawrence_danna]

include move version too

[labath]

I like where this is going.

lldb/source/Plugins/ScriptInterpreter/Python/PythonDataObjects.h

241–255

You can consider simplifying this further down to a "universal"/"sink" operator=(PythonObject other). Since the object is really just a pointer, the extra object being created won't hurt (in fact, the removal of &-indirection might make things faster).

[lawrence_danna]

lldb/source/Plugins/ScriptInterpreter/Python/PythonDataObjects.h

241–255

wouldn't that result in an extra retain and release every time a PythonObject was copied instead of referenced or moved?

[labath]

lldb/source/Plugins/ScriptInterpreter/Python/PythonDataObjects.h

241–255

No, it shouldn't, because the temporary PythonObject will be move-constructed (== no refcount traffic), if the operator= is called with an xvalue (if the rhs was not an xvalue, then you wouldn't end up calling the && overload anyway). Then you can move the temporary object into *this, and avoid refcount traffic again.

So, there is an additional PythonObject created, but it's move-constructed if possible, which should be efficient, if I understand these classes correctly. This is the recommended practice (at least by some) when you don't want to squeeze every last nanosecond of performance..

[lawrence_danna]

lldb/source/Plugins/ScriptInterpreter/Python/PythonDataObjects.h

241–255

How do you move the temporary object into *this, if you only have operator=(PythonObject other) to assign with?

[labath]

lldb/source/Plugins/ScriptInterpreter/Python/PythonDataObjects.h

241–255

In case that wasn't clear, the idea is to replace two operator= overloads with a single universal one taking a temporary. The advantage of that is less opportunities to implement move/copy incorrectly. The cost is one temporary move-constructed object more.

[lawrence_danna]

lldb/source/Plugins/ScriptInterpreter/Python/PythonDataObjects.h

241–255

so does that amount to just deleting the copy-assign, and keeping the move-assign how it is?

[labath]

lldb/source/Plugins/ScriptInterpreter/Python/PythonDataObjects.h

241–255

How do you move the temporary object into *this, if you only have operator=(PythonObject other) to assign with?

You need to do it manually, like the current && overload does, but you don't also need to implement the copy semantics in the const& overload.

so does that amount to just deleting the copy-assign, and keeping the move-assign how it is?

Almost. The implementation of move-assign would remain the same, but you'd drop the && (otherwise you'd lose the ability to copy-assign) from the signature. I.e.,

PythonObject &operator=(PythonObject other) {
    Reset();
    m_py_obj = std::exchange(other.m_py_obj, nullptr); // I just learned of this today so I have to show off.
    return *this;
  }

[lawrence_danna]

lldb/source/Plugins/ScriptInterpreter/Python/PythonDataObjects.h

241–255

oooooooh, i get it. It didn't occur to me that you could treat other as a rvalue without passing it as one in the parameter list, but of course you can because if it's pass-by-value then it's just a local variable by the time operator= gets its hands on it.

[lawrence_danna]

universal assignment

[labath]

lldb/source/Plugins/ScriptInterpreter/Python/PythonDataObjects.h

241–255

Yeah, isn't c++ fun? :)

I actually think this is one of the best features of c++11, as it allows you to write a single function that takes ownership of something more-or-less efficiently without any superfluous overloads or fancy &&s. The only problem is overcoming the learned "wisdom" that const & is the most efficient way to pass objects around.

lldb/source/Plugins/ScriptInterpreter/Python/PythonDataObjects.h

 class PythonObject {
 public:
   PythonObject() : m_py_obj(nullptr) {}
 
   PythonObject(PyRefType type, PyObject *py_obj) : m_py_obj(nullptr) {
     Reset(type, py_obj);
   }
 
-  PythonObject(const PythonObject &rhs) : m_py_obj(nullptr) { Reset(rhs); }
+  PythonObject(const PythonObject &rhs)
+      : PythonObject(PyRefType::Borrowed, rhs.m_py_obj) {}
 
   PythonObject(PythonObject &&rhs) {
     m_py_obj = rhs.m_py_obj;
     rhs.m_py_obj = nullptr;
   }
 
   ~PythonObject() { Reset(); }
 
   void Reset() {
     if (m_py_obj && Py_IsInitialized())
       Py_DECREF(m_py_obj);
     m_py_obj = nullptr;
   }
 
-  void Reset(const PythonObject &rhs) {
-    if (!rhs.IsValid())
-      Reset();
-    else
-      Reset(PyRefType::Borrowed, rhs.m_py_obj);
-  }
-
   // PythonObject is implicitly convertible to PyObject *, which will call the
   // wrong overload.  We want to explicitly disallow this, since a PyObject
   // *always* owns its reference.  Therefore the overload which takes a
   // PyRefType doesn't make sense, and the copy constructor should be used.
   void Reset(PyRefType type, const PythonObject &ref) = delete;
 
   void Reset(PyRefType type, PyObject *py_obj) {
     if (py_obj == m_py_obj)
public:
   PyObject *get() const { return m_py_obj; }
 
   PyObject *release() {
     PyObject *result = m_py_obj;
     m_py_obj = nullptr;
     return result;
   }
 
   PythonObject &operator=(const PythonObject &other) {
     Reset(PyRefType::Borrowed, other.get());
     return *this;
   }
 
   void Reset(PythonObject &&other) {
     Reset();
     m_py_obj = other.m_py_obj;
     other.m_py_obj = nullptr;
   }
 
   PythonObject &operator=(PythonObject &&other) {
     Reset(std::move(other));
     return *this;
   }
 
   PyObjectType GetObjectType() const;
 
   PythonString Repr() const;
 
   PythonString Str() const;
 
   static PythonObject ResolveNameWithDictionary(llvm::StringRef name,