I don't seem to get why would you use the move assignment operator
:
CLASSA & operator=(CLASSA && other); //move assignment operator
over, the copy assignment operator
:
CLASSA & operator=(CLASSA other); //copy assignment operator
The move assignment operator
takes an r-value reference
only e.g.
CLASSA a1, a2, a3;
a1 = a2 + a3;
In the copy assignment operator
, other
can be constructor using a copy constructor
or a move constructor
(if other
is initialized with an rvalue, it could be move-constructed --if move-constructor
defined--).
If it is copy-constructed
, we will be doing 1 copy and that copy can't be avoided.
If it is move-constructed
then the performance/behavior is identical to the one produced by the first overload.
My questions are:
1- Why would one want to implement the move assignment operator
.
2- If other
is constructed from an r-value then which assignment operator
would the compiler choose to call? And why?
You are not comparing like-with-like
If you are writing a move-only type like std::unique_ptr
then a move assignment operator would be your only choice.
The more typical case is where you have a copyable type in which case I think you have three options.
T& operator=(T const&)
T& operator=(T const&)
and T& operator=(T&&)
T& operator=(T)
and moveNote that having both the overloads you suggested in one class is not an option as it would be ambiguous.
Option 1 is the traditional C++98 option and will perform fine in most cases. However, if you need to optimize for r-values you could consider Option 2 and add a move assignment operator.
It is tempting to consider Option 3 and pass-by-value and then move which I think is what you are suggesting. In that case you only have to write one assignment operator. It accepts l-values and at the cost of only one extra move accepts r-values and many people will advocate this approach.
However, Herb Sutter pointed out in his "Back to the Basics! Essentials of Modern C++ Style" talk at CppCon 2014 that this option is problematic and can be much slower. In the case of l-values it will perform an unconditional copy and will not reuse any existing capacity. He provides numbers to backup his claims. The only exception is constructors where there is no existing capacity to reuse and you often have many parameters so pass by-value can reduce the number of overloads needed.
So I would suggest you start with Option 1 and move to Option 2 if you need to optimize for r-values.
Clearly, the two overloads are not equivalent:
T const&
would be needed for copyable types. Of course, for move-only types, like std::unique_ptr<T>
, defining this assignment operator is the appropriate choice.swap()
to replace the object's state with the state from the right-hand side. It has the advantage that the copy/move construction of the argument can often be elided.In any case, you wouldn't want to have both overloads in one class! When assigning from an lvalue, obviously, the version taking a value would be chosen (the other option isn't viable). However, both assignment operators are viable when assigning an rvalue, i.e., there would be an ambiguity. This can easily be verified by trying to compile this code:
struct foo
{
void operator=(foo&&) {}
void operator=(foo) {}
};
int main()
{
foo f;
f = foo();
}
To deal with a move- and copy construction separately you could define a pair of assignment operators using T&&
and T const&
as arguments. However, this results in having to implement two versions of essentially the same copy assignment while having just a T
as argument requires just one copy assignment to be implemented.
Thus, there are two obvious choices:
T::operator= (T&&)
.T::operator=(T)
.If you love us? You can donate to us via Paypal or buy me a coffee so we can maintain and grow! Thank you!
Donate Us With