Re: auto-generated move assignment and base or member with a by-value assignment

=?ISO-8859-2?Q?Krzysztof_Czai=F1ski?= <>
Wed, 26 Feb 2014 05:42:14 -0800 (PST)
On Monday, February 24, 2014 10:38:12 PM UTC+1, Daniel Kr?gler wrote:

Am 24.02.2014 15:23, schrieb Krzysztof Czainski:

On Monday, February 17, 2014 9:35:29 PM UTC+1, Daniel Krugler wrote:

Krzysztof Czainski:

So the standard treats A's assignment operator taking A by value as

Yes, this needs to be the case, because a operator= overload of the


T& T::operator=(T)

was already a copy-assignment operator in previous versions of C++.

But there were no move assignment operators, nor move constructors in
previous C++.


I intend it to be a universal assignment operator. Typically:

T() = default;
T( T const& ) = default;
T( T&& ) = default;
T& T::operator=( T x ) { swap(*this,x); return *this; }

Here, I see T& T::operator=( T ) as a replacement for:
T& T::operator=( T const& )
T& T::operator=( T&& )

Notice, that the following sets of overloads are ambiguous:

T& T::operator=( T )
T& T::operator=( T&& )


T& T::operator=( T const& )
T& T::operator=( T )

Therefore I believe

T& T::operator=( T )

is both the copy and move assignment operator.

Sure, it is possible to consider it conceptually this way, but
technically it is a copy assignment operator (as it had been in C++03)
and this difference becomes obvious, if you consider the effects on
"container" types - that is types, that either contain T as non-static
data member or as a direct base class: Because T& T::operator=(T)
declares a copy-assignment operator, the corresponding "container" type
will (implicitly) always declare a copy-assignment operator as well.

Yes, so with T& T::operator=( T ) I think both: copy and move asignments
in "container" classes should be generated. And with the fix by CWG 1402
they will be.

But then, calling:
T& T::operator=( T )
"copy assignment" is misleading to me, because it seems no more a copy
assignment than a move assignment.

I would argue that describing this function as copy assignment character
of T makes sense, because you can apply an lvalue of T as argument, but
for any move-assignment operator this would not be possible. And if the
argument is an lvalue, this function will never modify it, so it truly
is non-mutating for lvalues. But I understand the point that you are
trying to make: For non-constant rvalues this function behaves
potentially mutating and has therefore some aspects of a typical move

HTH & Greetings from Bremen,
- Daniel Kr??gler

Thank you for explaining this, Daniel, but this did not convince me.
I will try to support my statement, that calling T& T::operator=( T ) is
misleading. Consider these classes:

struct A // noncopyable
  A() = default;
  A( A const& ) = delete;
  A( A&& ) = delete; // this is redundant, here, isn't it?
  A& operator=( A ) {}

An object of type A can't be copy assigned to or move assigned to,
and yet in the same time it has a public "copy assignment operator".
Come on, in practice A isn't copy assignable, nor move assignable!

struct B // movable only
  B() = default;
  B( B const& ) = delete;
  B( B&& ) = default;
  B& operator=( B ) {}

An object of type B can't be copy assigned to, but can be move assigned to,
and yet in the same time it has a public "copy assignment operator".
Come on, in practice B isn't copy assignable, but it is move assignable!

struct C // copyable only
  C() = default;
  C( C const& ) = default;
  // C( C&& ) not declared, so that the copy ctor is used instead
  C& operator=( C ) {}

An object of type C can be copy assigned to and move assigned to,
and both will do the same -- copy. This is the case, where it in deed
it has a copy assignment operator.

struct D // copyable and movable
  D() = default;
  D( D const& ) = default;
  D( C&& ) = default;
  D& operator=( D ) {}

An object of type D can be copy assigned to and truly move assigned to.
I mean truly move assigned, no copy occurs. But it has a public
"copy assignment operator". Come on, in practice D is both copy assignable
and move assignable.

I wonder, what should the traits is_copy_assignable and is_move_assignable
report for the above four classes?

As shown above, a class that has a "copy assignment operator" in practice
behaves like it had other set of assignment operators (except for class C).
I call that misleading. I think it should be clearly stated, that
any class that has T& operator=( T ) has:
- a copy assignment operator iff it is copy constructable, and
- a move assignment operator iff it is move constructable.
(iff == if and only if)

I hope this shows my concern better.


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