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:
> > > > > > 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;
}
```
Repository:
rG LLVM Github Monorepo
CHANGES SINCE LAST ACTION
https://reviews.llvm.org/D69080/new/
https://reviews.llvm.org/D69080
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