[libcamera-devel] [PATCH] py: Drop redundant std::move()

[Tomi Valkeinen]

On 28/04/2023 17:51, Laurent Pinchart wrote:
> Hi Tomi,
> 
> On Wed, Apr 05, 2023 at 09:49:45AM +0300, Tomi Valkeinen wrote:
>> On 25/01/2023 11:42, Laurent Pinchart via libcamera-devel wrote:
>>> On Wed, Jan 25, 2023 at 12:09:40AM +0000, Kieran Bingham wrote:
>>>> Quoting Laurent Pinchart via libcamera-devel (2023-01-24 23:36:24)
>>>>> gcc-13 warns that the valueOrTuple() function has a redundant
>>>>> std::move() in a return statement:
>>>>>
>>>>> ../../src/py/libcamera/py_helpers.cpp: In instantiation of ‘pybind11::object valueOrTuple(const libcamera::ControlValue&) [with T = bool]’:
>>>>> ../../src/py/libcamera/py_helpers.cpp:38:28:   required from here
>>>>> ../../src/py/libcamera/py_helpers.cpp:28:35: error: redundant move in return statement [-Werror=redundant-move]
>>>>>      28 |                 return std::move(t);
>>>>
>>>> ohhh - this may be just too pedantic for me. Explicitly stating
>>>> std::move(t) when the compiler knows it is a move may be redundant to
>>>> the compiler, but it's not redundant to the reader?!
>>>>
>>>> Doesn't this help make it clear that the t is being moved... in which
>>>> case it's helpful self documenting code?
>>>>
>>>> I'm normally all for warnings, but this one is annoying.
>>>>
>>>> https://developers.redhat.com/blog/2019/04/12/understanding-when-not-to-stdmove-in-c
>>>> states that this isn't a 'pessimizing' operation, it's just redundant,
>>>> but it does make it clearer that a move is expected to occur?
>>>
>>> Adding more context, the function is implemented as
>>>
>>> 	if (cv.isArray()) {
>>> 		const T *v = reinterpret_cast<const T *>(cv.data().data());
>>> 		auto t = py::tuple(cv.numElements());
>>>
>>> 		for (size_t i = 0; i < cv.numElements(); ++i)
>>> 			t[i] = v[i];
>>>
>>> 		return std::move(t);
>>> 	}
>>>
>>> 	return py::cast(cv.get<T>());
>>>
>>> The type of 't' is py::tuple (replacing 'auto' with 'py::tuple' still
>>> produces the same warning), which inherits from py::object. We thus seem
>>> to be in the last case described by the above link:
>>>
>>>       There are situations where returning std::move (expr) makes sense,
>>>       however. The rules for the implicit move require that the selected
>>>       constructor take an rvalue reference to the returned object's type.
>>>       Sometimes that isn't the case. For example, when a function returns
>>>       an object whose type is a class derived from the class type the
>>>       function returns. In that case, overload resolution is performed a
>>>       second time, this time treating the object as an lvalue:
>>>
>>>       struct U { };
>>>       struct T : U { };
>>>
>>>       U f()
>>>       {
>>>               T t;
>>>               return std::move (t);
>>>       }
>>>
>>> g++-13 produces a warning when compiling that code:
>>>
>>> move.cpp: In function ‘U f()’:
>>> move.cpp:9:26: warning: redundant move in return statement [-Wredundant-move]
>>>       9 |         return std::move (t);
>>>         |                ~~~~~~~~~~^~~
>>> move.cpp:9:26: note: remove ‘std::move’ call
>>>
>>> This may also be a false positive of gcc ?
>>
>> I don't have gcc 13, nor does godbolt.org, but other gcc nor clang
>> versions don't complain.
>>
>> With some testing on godbolt, with and without std::move the end result
>> is the same (with -O2) on the compilers I tested.
>>
>> So... I don't know. The text you pasted seems to suggest that
>> std::move() would be needed there, but I don't see a diff (then again,
>> my test code is just test code, not the actual py code we have). I'm
>> fine either way, but if gcc 13 is not much used yet, maybe we should
>> wait a bit?
>>
>> Also, a bit beside the point, I'm actually a bit surprised that
>>
>> U f()
>> {
>> 	return T();
>> }
>>
>> works without warnings (even if I add fields to U and T). It's silently
>> throwing away the T specific parts, only keeping the U parts.
> 
> I tried this test code:
> 
> --------
> #include <functional>
> #include <iostream>
> 
> struct Base {
> 	Base()
> 	{
> 		std::cout << "Base::Base()" << std::endl;
> 	}
> 
> 	Base(const Base &)
> 	{
> 		std::cout << "Base::Base(const Base &)" << std::endl;
> 	}
> 
> 	Base(Base &&)
> 	{
> 		std::cout << "Base::Base(Base &&)" << std::endl;
> 	}
> };
> 
> struct Derived : Base {
> };
> 
> Base f()
> {
> 	Derived t;
> 	return std::move(t);
> }
> 
> Base g()
> {
> 	Derived t;
> 	return t;
> }
> 
> int main()
> {
> 	f();
> 	g();
> 
> 	return 0;
> }
> --------
> 
> $ g++-12 -W -Wall -o move2 move2.cpp && ./move2
> Base::Base()
> Base::Base(Base &&)
> Base::Base()
> Base::Base(const Base &)
> 
> $ g++-13 -W -Wall -o move2 move2.cpp && ./move2
> move2.cpp: In function ‘Base f()’:
> move2.cpp:27:25: warning: redundant move in return statement [-Wredundant-move]
>     27 |         return std::move(t);
>        |                ~~~~~~~~~^~~
> move2.cpp:27:25: note: remove ‘std::move’ call
> Base::Base()
> Base::Base(Base &&)
> Base::Base()
> Base::Base(Base &&)
> 
> This is annoying. The move seems to be redundant indeed with g++-13, but
> dropping it results in the copy constructor being used with g++-12 and
> earlier.
> 

What about:

	Derived t;
	Base& b = t;
	return std::move(b);

  Tomi