Is there any case where a return of a RValue Reference (&&) is useful?
Solution 1
There are a few occasions when it is appropriate, but they are relatively rare. The case comes up in one example when you want to allow the client to move from a data member. For example:
template <class Iter>
class move_iterator
{
private:
Iter i_;
public:
...
value_type&& operator*() const {return std::move(*i_);}
...
};
Solution 2
This follows up on towi's comment. You never want to return references to local variables. But you might have this:
vector<N> operator+(const vector<N>& x1, const vector<N>& x2) { vector<N> x3 = x1; x3 += x2; return x3; }
vector<N>&& operator+(const vector<N>& x1, vector<N>&& x2) { x2 += x1; return std::move(x2); }
vector<N>&& operator+(vector<N>&& x1, const vector<N>& x2) { x1 += x2; return std::move(x1); }
vector<N>&& operator+(vector<N>&& x1, vector<N>&& x2) { x1 += x2; return std::move(x1); }
This should prevent any copies (and possible allocations) in all cases except where both parameters are lvalues.
Solution 3
No. Just return the value. Returning references in general is not at all dangerous- it's returning references to local variables which is dangerous. Returning an rvalue reference, however, is pretty worthless in almost all situations (I guess if you were writing std::move
or something).
Solution 4
You can return by reference if you are sure the referenced object will not go out of scope after the function exits, e.g. it's a global object's reference, or member function returning reference to class fields, etc.
This returning reference rule is just same to both lvalue and rvalue reference. The difference is how you want to use the returned reference. As I can see, returning by rvalue reference is rare. If you have function:
Type&& func();
You won't like such code:
Type&& ref_a = func();
because it effectively defines ref_a as Type& since named rvalue reference is an lvalue, and no actual move will be performed here. It's quite like:
const Type& ref_a = func();
except that the actual ref_a is a non-const lvalue reference.
And it's also not very useful even you directly pass func() to another function which takes a Type&& argument because it's still a named reference inside that function.
void anotherFunc(Type&& t) {
// t is a named reference
}
anotherFunc(func());
The relationship of func( ) and anotherFunc( ) is more like an "authorization" that func() agrees anotherFunc( ) might take ownership of (or you can say "steal") the returned object from func( ). But this agreement is very loose. A non-const lvalue reference can still be "stolen" by callers. Actually functions are rarely defined to take rvalue reference arguments. The most common case is that "anotherFunc" is a class name and anotherFunc( ) is actually a move constructor.
Related videos on Youtube
towi
Programmer, Algorithmicist, Programming Languagcist, Pythonist, C++icist, Photographer, Boardgamer.
Updated on November 04, 2020Comments
-
towi over 3 years
Is there a reason when a function should return a RValue Reference? A technique, or trick, or an idiom or pattern?
MyClass&& func( ... );
I am aware of the danger of returning references in general, but sometimes we do it anyway, don't we?
T& T::operator=(T)
is just one idiomatic example. But how aboutT&& func(...)
? Is there any general place where we would gain from doing that? Probably different when one writes library or API code, compared to just client code? -
towi about 13 yearsI think during the early design of the C++0x there was a time when it was suggested that things like the move-assign and
T&& operator+(const T&,T&&)
should return a&&
. But that is gone now, in the final draft. That's why I ask. -
towi about 13 yearsAn excellent example. The pattern is, you want client code to move something -- allowing him to "steal". Yes, of course.
-
Puppy about 13 yearsLast time I checked, it was undefined to access after moving from.
-
Howard Hinnant about 13 yearsIn general a moved from object, when used in the std::lib, must meet all of the requirements specified for whatever part of the std::lib it is using. std-defined types must additionally guarantee that their moved from state is valid. Clients can call any function with that object as long as there are no pre-conditions on its value for said function-call.
-
Howard Hinnant about 13 yearsFinally, in the example above, there are no moved-from objects. std::move doesn't move. It only casts to rvalue. It is up to the client to move (or not) from that rvalue. That client will only access a moved-from value if he dereferences the move_iterator twice, without intervening iterator traversal.
-
fredoverflow about 13 yearsWouldn't it be safer to use
value_type
instead ofvalue_type&&
as the return type? -
Howard Hinnant about 13 yearsYes, I think so. However in this case I think the added benefit outweighs the risk. move_iterator is often used in generic code to turn a copying algorithm into a moving one (e.g. a move-version of vector::insert). If you then supply a type with an expensive copy & move to the generic code, you've got an extra copy gratuitously added. I'm thinking array<int, N> for example. When move-inserting a bunch of these into vector, you don't want to accidentally introduce an extra copy. On the risk side,
const X& x = *i
is pretty rare. I don't think I've ever seen it. -
Howard Hinnant about 13 yearsOn second thought, no, it wouldn't be safer. Rationale: The danger in returning a reference is that if the reference refers to a temporary, then the temporary can destruct before the reference does, thus leading to a dangling reference. In this example the reference will (almost always) refer to an lvalue. Very few iterators return prvalues, and those that do should probably not be adapted by move_iterator. So in the normal use case, even if the client holds the reference beyond the sequence point, the referred-to object still exists after the sequence point.
-
sellibitze about 13 yearsWhile it's possible, this is generally frowned upon because this approach has its own issues besides saving temporaries. See stackoverflow.com/questions/6006527
-
wjl almost 13 yearsDon't those return calls need to use std::move()?
-
Clinton almost 13 years@wjl: Good question, but I don't think so. std::move works without using std::move. I think the cast to && does the trick here.
-
Dave Abrahams over 12 yearsWell, it's still a wee bit safer to return a value, you have to admit. However, that makes the move eager as opposed to optional (depending on what the client of
operator*
does with it) -
Potatoswatter about 11 yearsReturning by value or rvalue reference aren't comparable in risk, because they do completely different things. Returning by RR is exactly the same as adding
std::move
around the function call, that's all. And since it only makes sense to move an lvalue, use-cases of prvalues are just another story. -
M.M about 9 years@Clinton there's no cast in your code, you have to
return std::move(x2);
etc. Or you could write a cast to rvalue reference type, but that's just whatmove
does anyway. -
Aaron McDaid almost 9 years@Clinton, I've added in the
move
s. Matt is correct. I've tested it -
M.M almost 9 yearsThis code is a bad idea (not sure how I overlooked that in my previous comment) because it may return a dangling reference. For example, say you do
vector<N>&& x = vec1 + {1,2,3};
. The second argument will bind to the parametervector<N>&& x2
, however this temporary only lasts for the full-expression it was created in, leavingx
dangling. -
M.M almost 9 yearsThe code is only correct if the return value is either unused, or assigned to an object -- but then you may as well have returned by value and taken the arguments by value and let copy elision do its thing.
-
ThomasMcLeod almost 9 yearsIs this valid if
*i_
isconst
? Just asking. -
Howard Hinnant almost 9 years@ThomasMcLeod: If
value_type
is alsoconst
, it is valid, but there will be no moving, just a copy. Ifvalue_type
is notconst
, then you'll get a compile-time error complaining about assigning aconst
value to a non-const reference. -
TemplateRex over 8 yearsfor
vector<N>&& x2
, don't you wantx1 += std::move(x2);
? -
Emile Cormier about 8 yearsWouldn't this best be used with an r-value ref qualifier?
-
Howard Hinnant about 8 years@EmileCormier: No. The purpose of
move_iterator
is to turn generic "copy" algorithms into "move" algorithms. See open-std.org/jtc1/sc22/wg21/docs/papers/2005/n1771.html and search for "replace_copy".replace_copy
+move_iterator
allows you to buildreplace_move
for yourself. But thereplace_copy
algorithm need do nothing special. It is ignorant of the fact that it is dealing withmove_iterator
. It dereferences this iterator just as it would any other. -
Emile Cormier about 8 years@HowardHinnant, sorry I had tunnel vision when I posted that comment. I was only looking at the
operator*
line and didn't pay attention to themove_iterator
class around it.