[Lldb-commits] [PATCH] D69080: eliminate one form of PythonObject::Reset()

[Pavel Labath via Phabricator via lldb-commits]

labath added inline comments.


================
Comment at: lldb/source/Plugins/ScriptInterpreter/Python/PythonDataObjects.h:235-245
   PythonObject &operator=(const PythonObject &other) {
     Reset(PyRefType::Borrowed, other.get());
     return *this;
   }
 
-  void Reset(PythonObject &&other) {
+  PythonObject &operator=(PythonObject &&other) {
     Reset();
----------------
lawrence_danna wrote:
> labath wrote:
> > lawrence_danna wrote:
> > > labath wrote:
> > > > lawrence_danna wrote:
> > > > > labath wrote:
> > > > > > lawrence_danna wrote:
> > > > > > > labath wrote:
> > > > > > > > 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).
> > > > > > > wouldn't that result in an extra retain and release every time a PythonObject was copied instead of referenced or moved?
> > > > > > 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..
> > > > > How do you move the temporary object into *this, if you only have `operator=(PythonObject other)` to assign with?
> > > > 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.
> > > so does that amount to just deleting the copy-assign, and keeping the move-assign how it is?
> > > 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;
> >   }
> > ```
> 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.
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.


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  https://reviews.llvm.org/D69080/new/

https://reviews.llvm.org/D69080