I am trying to understand C++11 rvalue references and how to use them for optimal performance in my code.
Let's say we have a class A that has a member pointer to a large amount of dynamically allocated data.
Furthermore, a method foo(const A& a) that does something with an object of class A.
I want to prevent the copy constructor of A from being called when an object of A is passed to the function foo, since in that case it will perform a deep copy of the underlying heap data.
I tested passing an lvalue reference:
A a;
foo(a);
and passing an rvalue reference:
foo(A());
In both cases the copy constructor was not called.
Is this expected or is this due to some optimization of my compiler (Apple LLVM 5.1)? Is there any specification about this?
That is expected. If you pass an argument to a reference type parameter (whether lvalue or rvalue reference), the object will not be copied. That is the whole point of references.
The confusion you're having is pretty common. The choice of copy or move constructor only occurs when passing an object by value. For example:
void foo(A a);
When passing an A object to this function, the compiler will determine whether to use the copy or move constructor depending on whether the expression you pass is an lvalue or rvalue expression.
On the other hand, none of the following functions would even try to invoke the copy or move constructor because no object is being constructed:
void foo(A& a);
void foo(const A& a);
void foo(A&& a);
void foo(const A&& a);
It's important to note that you should rarely (if ever) have any reason to write a function, other than a move constructor/assignment operator, that takes an rvalue reference. You should be deciding between passing by value and passing by const lvalue reference:
If you're going to need a copy of the object inside the function anyway (perhaps because you want to modify a copy or pass it to another function), take it by value (A). This way, if you're given an lvalue, it'll have to be copied (you can't avoid this), but if you're given an rvalue, it'll be optimally moved into your function.
If you're not going to need a copy of the object, take it by const lvalue reference (const A&). This way, regardless of whether you're given an lvalue or rvalue, no copy will take place. You shouldn't use this when you do need to copy it though, because it prevents you from utilising move semantics.
From the sounds of it, you're not going to make any copies at all, so a const A& parameter would work.
Related
So the parameter list for a copy constructor consists of an const lvalue reference, like const B& x.
The parameter list for a move constructor, however, consists of an rvalue reference, like B&& x.
When I discovered this, it seemed odd to me, so I tried to define a function taking a const lvalue reference as an argument:
void calculator(const Intvec& veccor)
{
cout << "in veccor" << veccor.m_size << "\n";
}
I already knew that const lvalue reference can bind to anything, so the result was sort of expected. I tried calling the function with both and lvalue and an rvalue and as expected, it all worked:
calculator(Intvec(33)); //works
Intvec newvec(22);
calculator(newvec); //works
I then tried to change the parameter list of calculator to an rvalue reference, and as expected, only the rvalue worked when calling it.
So since a const lvalue reference can take both lvalues and rvalues as an argument, why does the move constructor not just use a const lvalue reference instead of an rvalue reference?
And why not just always use const lvalue references in general, like in cases where you would not do the same as what is happening in a move constructor?
If a move constructor accepted a const lvalue reference, then such declaration of a move constructor would be indistinguishable from a copy constructor. There has to be a way to distinguish them, and the language has been specified such that a move constructor takes an rvalue reference as the argument.
A move constructor that would accept a const lvalue reference would not allow the moved from object to be modified, so you couldn't do anything that you couldn't do in a copy constructor. In fact, such move constructor would be in every way identical to a copy constructor. Why would one call something a move constructor when it is exactly the same as a copy constructor?
PS. Your experiment reveals an interesting fact: As long as a class has a copy constructor (and the move constructor is not explicitly deleted), whether it has a move constructor or not does not affect how that object can be used. In any case where a move is appropriate, a copy can be used instead if the class has no move constructor.
Let's say I got a Foo class containing an std::vector constructed from std::unique_ptr objects of another class, Bar.
typedef std::unique_ptr<Bar> UniqueBar;
class Foo {
std::vector<UniqueBar> bars;
public:
void AddBar(UniqueBar&& bar);
};
void Foo::AddBar(UniqueBar&& bar) {
bars.push_back(bar);
}
This one results in a compilation error (in g++ 4.8.1) saying that the the copy constructor of std::unique_ptr is deleted, which is reasonable. The question here is, since the bar argument is already an rvalue reference, why does the copy constructor of std::unique_ptr is called instead of its move constructor?
If I explicitly call std::move in Foo::AddBar then the compilation issue goes away but I don't get why this is needed. I think it's quite redundant.
So, what am I missing?
Basically, every object which has a name is an lvalue. When you pass an object to a function using an rvalue reference the function actually sees an lvalue: it is named. What the rvalue reference does, however, indicate is that it came from an object which is ready to be transferred.
Put differently, rvalue references are assymmetrical:
they can only receive rvalues, i.e., either temporary objects, objects about to go away, or objects which look as if they are rvalues (e.g., the result of std::move(o))
the rvalue reference itself looks, however, like an lvalue
Confusing as it might seem, an rvalue-reference binds to an rvalue, but used as an expression is an lvalue.
bar is actually an lvalue, so you need to pass it through std::move, so that it is seen as an rvalue in the call to push_back.
The Foo::AddBar(UniqueBar&& bar) overload simply ensures that this overload is picked when an rvalue is passed in a call to Foo::AddBar. But the bar argument itself has a name and is an lvalue.
bar is defined as an rvalue-reference, but its value-category is an lvalue. This is so because the object has a name. If it has a name, it's an lvalue. Therefore an explicit std::move is necessary because the intention is to get rid of the name and return an xvalue (eXpiring-rvalue).
For example:
void f(T&& t); // probably making a copy of t
void g()
{
T t;
// do something with t
f(std::move(t));
// probably something else not using "t"
}
Is void f(T const& t) equivalent in this case because any good compiler will produce the same code? I'm interested in >= VC10 and >= GCC 4.6 if this matters.
EDIT:
Based on the answers, I'd like to elaborate the question a bit:
Comparing rvalue-reference and pass-by-value approaches, it's so easy to forgot to use std::move in pass-by-value. Can compiler still check that no more changes are made to the variable and eliminate an unnecessary copy?
rvalue-reference approach makes only optimized version "implicit", e.g. f(T()), and requires the user to explicitly specify other cases, like f(std::move(t)) or to explicitly make a copy f(T(t)); if the user isn't done with t instance. So, in this optimization-concerned light, is rvalue-reference approach considered good?
It's definitely not the same. For once T && can only bind to rvalues, while T const & can bind both to rvalues and to lvalues. Second, T const & does not permit any move optimizations. If you "probably want to make a copy of t", then T && allows you to actually make a move-copy of t, which is potentially more efficient.
Example:
void foo(std::string const & s) { std::string local(s); /* ... */ }
int main()
{
std::string a("hello");
foo(a);
}
In this code, the string buffer containing "hello" must exist twice, once in the body of main, and another time in the body of foo. By contrast, if you used rvalue references and std::move(a), the very same string buffer can be "moved around" and only needs to be allocated and populated one single time.
As #Alon points out, the right idiom is in fact passing-by-value:
void foo(std::string local) { /* same as above */ }
int main()
{
std::string a("hello");
foo(std::move(a));
}
Well, it depends what f does with t, if it creates a copy of it, then I would even go at length of doing this:
void f(T t) // probably making a copy of t
{
m_newT = std::move(t); // save it to a member or take the resources if it is a c'tor..
}
void g()
{
T t;
// do something with t
f(std::move(t));
// probably something else not using "t"
}
Then you allow the move c'tors optimization to happen, you take 't' resources in any case, and if it was 'moved' to your function, then you even gain the non copy of moving it to the function, and if it was not moved then you probably had to have one copy
Now if at later on in the code you'd have:
f(T());
Then ta da, free move optimization without the f user even knowing..
Note that quote: "is void f(T const& t) equivalent in this case because any good compiler will produce the same code?"
It is not equivelent, it is LESS work, because only the "pointer" is transferred and no c'tors are called at all, neither move nor anything else
Taking an const lvalue reference and taking an rvalue reference are two different things.
Similarities:
Neither will cause an copy or move to take place because they are both references. A reference just references an object, it doesn't copy/move it in any way.
Differences:
A const lvalue reference will bind to anything (lvalue or rvalue). An rvalue reference will only bind to non-const rvalues - much more limited.
The parameter inside the function cannot be modified when it is a const lvalue reference. It can be modified when it's an rvalue reference (since it is non-const).
Let's look at some examples:
Taking const lvalue reference: void f(const T& t);
Passing an lvalue:
T t; f(t);
Here, t is an lvalue expression because it's the name of the object. A const lvalue reference can bind to anything, so t will happily be passed by reference. Nothing is copied, nothing is moved.
Passing an rvalue:
f(T());
Here, T() is an rvalue expression because it creates a temporary object. Again, a const lvalue reference can bind to anything, so this is okay. Nothing is copied, nothing is moved.
In both of these cases, the t inside the function is a reference to the object passed in. It can't be modified by the reference is const.
Taking an rvalue reference: `void f(T&& t);
Passing an lvalue:
T t;
f(t);
This will give you a compiler error. An rvalue reference will not bind to an lvalue.
Passing an rvalue:
f(T());
This will be fine because an rvalue reference can bind to an rvalue. The reference t inside the function will refer to the temporary object created by T().
Now let's consider std::move. First things first: std::move doesn't actually move anything. The idea is that you give it an lvalue and it turns it into an rvalue. That's all it does. So now, if your f takes an rvalue reference, you could do:
T t;
f(std::move(t));
This works because, although t is an lvalue, std::move(t) is an rvalue. Now the rvalue reference can bind to it.
So why would you ever take an rvalue reference argument? In fact, you shouldn't need to do it very often, except for defining move constructors and assignment operators. Whenever you define a function that takes an rvalue reference, you almost certainly want to give a const lvalue reference overload. They should almost always come in pairs:
void f(const T&);
void f(T&&);
Why is this pair of functions useful? Well, the first will be called whenever you give it an lvalue (or a const rvalue) and the second will be called whenever you give it a modifiable rvalue. Receiving an rvalue usually means that you've been given a temporary object, which is great news because that means you can ravage its insides and perform optimizations based on the fact that you know it's not going to exist for much longer.
So having this pair of functions allows you to make an optimization when you know you're getting a temporary object.
There's a very common example of this pair of functions: the copy and move constructors. They are usually defined like so:
T::T(const T&); // Copy constructor
T::T(T&&); // Move constructor
So a move constructor is really just a copy constructor that is optimized for when receiving a temporary object.
Of course, the object being passed isn't always a temporary object. As we've shown above, you can use std::move to turn an lvalue into an rvalue. Then it appears to be a temporary object to the function. Using std::move basically says "I allow you to treat this object as a temporary object." Whether it actually gets moved from or not is irrelevant.
However, beyond writing copy constructors and move constructors, you'd better have a good reason for using this pair of functions. If you're writing a function that takes an object and will behave exactly the same with it regardless of whether its a temporary object or not, simply take that object by value! Consider:
void f(T t);
T t;
f(t);
f(T());
In the first call to f, we are passing an lvalue. That will be copied into the function. In the second call to f, we are passing an rvalue. That object will be moved into the function. See - we didn't even need to use rvalue references to cause the object to be moved efficiently. We just took it by value! Why? Because the constructor that is used to make the copy/move is chosen based on whether the expression is an lvalue or an rvalue. Just let the copy/move constructors do their job.
As to whether different argument types result in the same code - well that's a different question entirely. The compiler operates under the as-if rule. This simply means that as long as the program behaves as the standard dictates, the compiler can emit whatever code it likes. So the functions may emit the same code if they happen to do exactly the same thing. Or they may not. However, it's a bad sign if you're functions that take a const lvalue reference and an rvalue reference are doing the same thing.
In several places I've seen the recommended signatures of copy and move constructors given as:
struct T
{
T();
T(const T& other);
T(T&& other);
};
Where the copy constructor takes a const reference, and the move constructor takes a non-const rvalue reference.
As far as I can see though, this prevents me taking advantage of move semantics when returning const objects from a function, such as in the case below:
T generate_t()
{
const T t;
return t;
}
Testing this with VC11 Beta, T's copy constructor is called, and not the move constructor. Even using return std::move(t); the copy constructor is still called.
I can see how this makes sense, since t is const so shouldn't bind to T&&. Using const T&& in the move constructor signature works fine, and makes sense, but then you have the problem that because other is const, you can't null its members out if they need to be nulled out - it'll only work when all members are scalars or have move constructors with the right signature.
It looks like the only way to make sure the move constructor is called in the general case to have made t non-const in the first place, but I don't like doing that - consting things is good form and I wouldn't expect the client of T to know that they had to go against that form in order to increase performance.
So, I guess my question is twofold; first, should a move constructor take a const or non-const rvalue reference? And second: am I right in this line of reasoning? That I should stop returning things that are const?
It should be a non-const rvalue reference.
If an object is placed in read-only memory, you can't steal resources from it, even if its formal lifetime is ending shortly. Objects created as const in C++ are allowed to live in read-only memory (using const_cast to try to change them results in undefined behavior).
A move constructor should normally take a non-const reference.
If it were possible to move from a const object it would usually imply that it was as efficient to copy an object as it was to "move" from it. At this point there is normally no benefit to having a move constructor.
You are also correct that if you have a variable that you are potentially going to want to move from then it will need to be non-const.
As I understand it this is the reason that Scott Meyers has changed his advice on returning objects of class type by value from functions for C++11. Returning objects by const qualified value does prevent unintentionally modification of a temporary object but it also inhibits moving from the return value.
Should a move constructor take a const or non-const rvalue reference?
It should take non-const rvalue reference. The rvalue references first of all don't make sense in their const forms simply because you want to modify them (in a way, you want to "move" them, you want their internals for yourself ).
Also, they have been designed to be used without const and I believe the only use for a const rvalue reference is something very very arcane that Scott Meyers mentioned in this talk (from the time 42:20 to 44:47).
Am I right in this line of reasoning? That I should stop returning things that are const?
This is a bit of too general question to answer I reckon. In this context, I think it's worth mentioning that there's std::forward functionality that will preserve both rvalue-ness and lvalue-ness as well as const-ness and it will also avoid creating a temporary as a normal function would do should you return anything passed to it.
This returning would also cause the rvalue reference to be "mangled" into lvalue reference and you generally don't want that, hence, perfect forwarding with the aforementioned functionality solves the issue.
That being said, I suggest you simply take a look at the talk that I posted a link to.
In addition to what is said in other answers, sometimes there are reasons for a move constructor or a function to accept a const T&&. For example, if you pass the result of a function that returns a const object by value to a constructor, T(const T&) will be called instead of T(T&&) as one would probably expect (see function g below).
This is the reason behind deleting overloads that accept constT&& for std::ref and std::cref instead of those that accept T&&.
Specifically, the order of preference during overload resolution is as follows:
struct s {};
void f ( s&); // #1
void f (const s&); // #2
void f ( s&&); // #3
void f (const s&&); // #4
const s g ();
s x;
const s cx;
f (s ()); // rvalue #3, #4, #2
f (g ()); // const rvalue #4, #2
f (x); // lvalue #1, #2
f (cx); // const lvalue #2
See this article for more details.
When you have a derived object with a move constructor, and the base object also has move semantics, what is the proper way to call the base object move constructor from the derived object move constructor?
I tried the most obvious thing first:
Derived(Derived&& rval) : Base(rval)
{ }
However, this seems to end up calling the Base object's copy constructor. Then I tried explicitly using std::move here, like this:
Derived(Derived&& rval) : Base(std::move(rval))
{ }
This worked, but I'm confused why it's necessary. I thought std::move merely returns an rvalue reference. But since in this example rval is already an rvalue reference, the call to std::move should be superfluous. But if I don't use std::move here, it just calls the copy constructor. So why is the call to std::move necessary?
rval is not a Rvalue. It is an Lvalue inside the body of the move constructor. That's why we have to explicitly invoke std::move.
Refer this. The important note is
Note above that the argument x is
treated as an lvalue internal to the
move functions, even though it is
declared as an rvalue reference
parameter. That's why it is necessary
to say move(x) instead of just x when
passing down to the base class. This
is a key safety feature of move
semantics designed to prevent
accidently moving twice from some
named variable. All moves occur only
from rvalues, or with an explicit cast
to rvalue such as using std::move. If
you have a name for the variable, it
is an lvalue.
Named R-value references are treated as L-value.
So we need std::move to convert it to R-Value.
You really should use std::forward(obj) rather than std::move(obj). Forward will return the proper rvalue or lvalue based on the what obj is whereas move will turn an lvalue into an rvalue.