I think there's something I'm not quite understanding about rvalue references. Why does the following fail to compile (VS2012) with the error 'foo' : cannot convert parameter 1 from 'int' to 'int &&'?
void foo(int &&) {}
void bar(int &&x) { foo(x); };
I would have assumed that the type int && would be preserved when passed from bar into foo. Why does it get transformed into int once inside the function body?
I know the answer is to use std::forward:
void bar(int &&x) { foo(std::forward<int>(x)); }
so maybe I just don't have a clear grasp on why. (Also, why not std::move?)
I always remember lvalue as a value that has a name or can be addressed. Since x has a name, it is passed as an lvalue. The purpose of reference to rvalue is to allow the function to completely clobber value in any way it sees fit. If we pass x by reference as in your example, then we have no way of knowing if is safe to do this:
void foo(int &&) {}
void bar(int &&x) {
foo(x);
x.DoSomething(); // what could x be?
};
Doing foo(std::move(x)); is explicitly telling the compiler that you are done with x and no longer need it. Without that move, bad things could happen to existing code. The std::move is a safeguard.
std::forward is used for perfect forwarding in templates.
Why does it get transformed into int once inside the function body?
It doesn't; it's still a reference to an rvalue.
When a name appears in an expression, it's an lvalue - even if it happens to be a reference to an rvalue. It can be converted into an rvalue if the expression requires that (i.e. if its value is needed); but it can't be bound to an rvalue reference.
So as you say, in order to bind it to another rvalue reference, you have to explicitly convert it to an unnamed rvalue. std::forward and std::move are convenient ways to do that.
Also, why not std::move?
Why not indeed? That would make more sense than std::forward, which is intended for templates that don't know whether the argument is a reference.
It's the "no name rule". Inside bar, x has a name ... x. So it's now an lvalue. Passing something to a function as an rvalue reference doesn't make it an rvalue inside the function.
If you don't see why it must be this way, ask yourself -- what is x after foo returns? (Remember, foo is free to move x.)
rvalue and lvalue are categories of expressions.
rvalue reference and lvalue reference are categories of references.
Inside a declaration, T x&& = <initializer expression>, the variable x has type T&&, and it can be bound to an expression (the ) which is an rvalue expression. Thus, T&& has been named rvalue reference type, because it refers to an rvalue expression.
Inside a declaration, T x& = <initializer expression>, the variable x has type T&, and it can be bound to an expression (the ) which is an lvalue expression (++). Thus, T& has been named lvalue reference type, because it can refer to an lvalue expression.
It is important then, in C++, to make a difference between the naming of an entity, that appears inside a declaration, and when this name appears inside an expression.
When a name appears inside an expression as in foo(x), the name x alone is an expression, called an id-expression. By definition, and id-expression is always an lvalue expression and an lvalue expressions can not be bound to an rvalue reference.
When talking about rvalue references it's important to distinguish between two key unrelated steps in the lifetime of a reference - binding and value semantics.
Binding here refers to the exact way a value is matched to the parameter type when calling a function.
For example, if you have the function overloads:
void foo(int a) {}
void foo(int&& a) {}
Then when calling foo(x), the act of selecting the proper overload involves binding the value x to the parameter of foo.
rvalue references are only about binding semantics.
Inside the bodies of both foo functions the variable a acts as a regular lvalue. That is, if we rewrite the second function like this:
void foo(int&& a) {
foo(a);
}
then intuitively this should result in a stack overflow. But it doesn't - rvalue references are all about binding and never about value semantics. Since a is a regular lvalue inside the function body, then the first overload foo(int) will be called at that point and no stack overflow occurs. A stack overflow would only occur if we explicitly change the value type of a, e.g. by using std::move:
void foo(int&& a) {
foo(std::move(a));
}
At this point a stack overflow will occur because of the changed value semantics.
This is in my opinion the most confusing feature of rvalue references - that the type works differently during and after binding. It's an rvalue reference when binding but it acts like an lvalue reference after that. In all respects a variable of type rvalue reference acts like a variable of type lvalue reference after binding is done.
The only difference between an lvalue and an rvalue reference comes when binding - if there is both an lvalue and rvalue overload available, then temporary objects (or rather xvalues - eXpiring values) will be preferentially bound to rvalue references:
void goo(const int& x) {}
void goo(int&& x) {}
goo(5); // this will call goo(int&&) because 5 is an xvalue
That's the only difference. Technically there is nothing stopping you from using rvalue references like lvalue references, other than convention:
void doit(int&& x) {
x = 123;
}
int a;
doit(std::move(a));
std::cout << a; // totally valid, prints 123, but please don't do that
And the keyword here is "convention". Since rvalue references preferentially bind to temporary objects, then it's reasonable to assume that you can gut the temporary object, i.e. move away all of its data away from it, because after the call it's not accessible in any way and is going to be destroyed anyway:
std::vector<std::string> strings;
string.push_back(std::string("abc"));
In the above snippet the temporary object std::string("abc") cannot be used in any way after the statement in which it appears, because it's not bound to any variable. Therefore push_back is allowed to move away its contents instead of copying it and therefore save an extra allocation and deallocation.
That is, unless you use std::move:
std::vector<std::string> strings;
std::string mystr("abc");
string.push_back(std::move(mystr));
Now the object mystr is still accessible after the call to push_back, but push_back doesn't know this - it's still assuming that it's allowed to gut the object, because it's passed in as an rvalue reference. This is why the behavior of std::move() is one of convention and also why std::move() by itself doesn't actually do anything - in particular it doesn't do any movement. It just marks its argument as "ready to get gutted".
The final point is: rvalue references are only useful when used in tandem with lvalue references. There is no case where an rvalue argument is useful by itself (exaggerating here).
Say you have a function accepting a string:
void foo(std::string);
If the function is going to simply inspect the string and not make a copy of it, then use const&:
void foo(const std::string&);
This always avoids a copy when calling the function.
If the function is going to modify or store a copy of the string, then use pass-by-value:
void foo(std::string s);
In this case you'll receive a copy if the caller passes an lvalue and temporary objects will be constructed in-place, avoiding a copy. Then use std::move(s) if you want to store the value of s, e.g. in a member variable. Note that this will work efficiently even if the caller passes an rvalue reference, that is foo(std::move(mystring)); because std::string provides a move constructor.
Using an rvalue here is a poor choice:
void foo(std::string&&)
because it places the burden of preparing the object on the caller. In particular if the caller wants to pass a copy of a string to this function, they have to do that explicitly;
std::string s;
foo(s); // XXX: doesn't compile
foo(std::string(s)); // have to create copy manually
And if you want to pass a mutable reference to a variable, just use a regular lvalue reference:
void foo(std::string&);
Using rvalue references in this case is technically possible, but semantically improper and totally confusing.
The only, only place where an rvalue reference makes sense is in a move constructor or move assignment operator. In any other situation pass-by-value or lvalue references are usually the right choice and avoid a lot of confusion.
Note: do not confuse rvalue references with forwarding references that look exactly the same but work totally differently, as in:
template <class T>
void foo(T&& t) {
}
In the above example t looks like a rvalue reference parameter, but is actually a forwarding reference (because of the template type), which is an entirely different can of worms.
Related
Consider the following code: (compile)
void f(int&&) {}
template<class T>
void s(T&& value)
{
using rv_ref = decay_t<T>&&;
static_assert( is_same_v<decltype(value), rv_ref> );
f(static_cast<rv_ref>(value)); // ok, but we've checked `value` is already of this type
//f(value); // error: no matching function
}
int main()
{
s(9);
}
What makes me flinch is the fact that even though the type of value is int&&, when it's being used as a value [and not an expression(as with decltype)], it suddenly becomes an l-value.
How is this explained to make sense?
You write this:
even though the type of value is int&&, when it's being used as a value [and not an expression(as with decltype)], it suddenly becomes an l-value.
One high-level way to think about lvalues and rvalues is that lvalues have names and can be assigned to. value obviously has a name, so it shouldn't be surprising that it's an lvalue.
One way to think of std::move() or your static cast is that it not only casts its argument to the correct type, but it also produces an rvalue expression.
Thinking about your two functions s and f:
Presumably, your function s takes an rvalue reference because you want to operate on rvalues, e.g. steal their resources with a move operation.
In such a function, you might want to call functions on value that either (i) steal its resources, or (ii) treat it as an lvalue and do not steal its resources.
The language lets you call std::move() for case (i), to tell f that f may steal resources.
The language lets you pass value to functions without std::move(), as an lvalue, for case (ii), so you can call functions without worrying about that. This might be desirable if you want s to steal resources later on.
This question is pretty similar: Rvalue Reference is Treated as an Lvalue?
The name of a variable used as an expression is always an lvalue, even if the variable's type is an rvalue reference. (Though decltype has some additional rules, not just looking at the expression's value category.) This is an intentional choice of the C++ language, because otherwise it would be too easy to accidentally move from a variable.
For example:
void debug_str_val(std::string var_name, std::string value);
void MyClass::set_name(const std::string& name)
{
debug_str_val("name", name);
m_name = name;
}
void MyClass::set_name(std::string&& name)
{
debug_str_val("name", name); // *
m_name = std::move(name);
}
On the line marked with a *, name is treated as an lvalue, so debug_str_val gets a copy of the value. (It maybe could use const std::string& parameters instead, but it doesn't.) If name were treated as an rvalue, the debug_str_val would be moved from, and then an unpredictable value would be assigned to m_name.
Once an object has a name, it could be used more than once, even if that name is an rvalue reference. So C++ uses it as an lvalue by default and lets the programmer say when they want it to be used as an rvalue, using std::move, or sometimes std::forward, or an explicit cast to rvalue reference type.
I was wondering about a c++ behaviour when an r-value is passed among functions.
Look at this simple code:
#include <string>
void foo(std::string&& str) {
// Accept a rvalue of str
}
void bar(std::string&& str) {
// foo(str); // Does not compile. Compiler says cannot bind lvalue into rvalue.
foo(std::move(str)); // It feels like a re-casting into a r-value?
}
int main(int argc, char *argv[]) {
bar(std::string("c++_rvalue"));
return 0;
}
I know when I'm inside bar function I need to use move function in order to invoke foo function. My question now is why?
When I'm inside the bar function the variable str should already be an r-value, but the compiler acts like it is a l-value.
Can somebody quote some reference to the standard about this behaviour?
Thanks!
str is a rvalue reference, i.e. it is a reference only to rvalues. But it is still a reference, which is a lvalue. You can use str as a variable, which also implies that it is an lvalue, not a temporary rvalue.
An lvalue is, according to §3.10.1.1:
An lvalue (so called, historically, because lvalues could appear on the left-hand side of an assignment expression) designates a function or an object. [ Example: If E is an expression of pointer type, then *E is an lvalue expression referring to the object or function to which E points. As another example, the result of calling a function whose return type is an lvalue reference is an lvalue. —end example ]
And an rvalue is, according to §3.10.1.4:
An rvalue (so called, historically, because rvalues could appear on the right-hand side of an assignment
expression) is an xvalue, a temporary object (12.2) or subobject thereof, or a value that is not associated with an object.
Based on this, str is not a temporary object, and it is associated with an object (with the object called str), and so it is not an rvalue.
The example for the lvalue uses a pointer, but it is the same thing for references, and naturally for rvalue references (which are only a special type of references).
So, in your example, str is an lvalue, so you have to std::move it to call foo (which only accepts rvalues, not lvalues).
The "rvalue" in "rvalue reference" refers to the kind of value that the reference can bind to:
lvalue references can bind to lvalues
rvalue references can bind to rvalues
(+ a bit more)
That's all there's to it. Importantly, it does not refer to the value that get when you use the reference. Once you have a reference variable (any kind of reference!), the id-expression naming that variable is always an lvalue. Rvalues occur in the wild only as either temporary values, or as the values of function call expressions, or as the value of a cast expression, or as the result of decay or of this.
There's a certain analogy here with dereferencing a pointer: dereferencing a pointer is always an lvalue, no matter how that pointer was obtained: *p, *(p + 1), *f() are all lvalues. It doesn't matter how you came by the thing; once you have it, it's an lvalue.
Stepping back a bit, maybe the most interesting aspect of all this is that rvalue references are a mechanism to convert an rvalue into an lvalue. No such mechanism had existed prior to C++11 that produced mutable lvalues. While lvalue-to-rvalue conversion has been part of the language since its very beginnings, it took much longer to discover the need for rvalue-to-lvalue conversion.
My question now is why?
I'm adding another answer because I want to emphasize an answer to the "why".
Even though named rvalue references can bind to an rvalue, they are treated as lvalues when used. For example:
struct A {};
void h(const A&);
void h(A&&);
void g(const A&);
void g(A&&);
void f(A&& a)
{
g(a); // calls g(const A&)
h(a); // calls h(const A&)
}
Although an rvalue can bind to the a parameter of f(), once bound, a is now treated as an lvalue. In particular, calls to the overloaded functions g() and h() resolve to the const A& (lvalue) overloads. Treating a as an rvalue within f would lead to error prone code: First the "move version" of g() would be called, which would likely pilfer a, and then the pilfered a would be sent to the move overload of h().
Reference.
So I have the following function:
void scan(std::istream& is, Handler& h);
I want to call it in different ways, like:
scan(std::cin, Handler());
scan(std::ifstream("myfile"), myhandler);
The compiler complains about std::ifstream("myfile") and Handler() of being rvalues being passed as non-const references, so the complaint is legitimate, but what can I do?
Neither function parameters cannot be const (istream is modified while read and the handler changes its state during callbacks).
If I change the parameter types to rvalue references (&&) then I will not be able to pass std::cin and sometimes I really care about the final state of myhandler thus I cannot apply std::move on them neither.
In principle I could make the parameters as universal references via template or auto&& type deduction and thus overload this function for all possible combinations of lvalue and rvalue references, but I have no intention of overloading this function for other types than I have already specified.
Are there any other options?
Somehow this whole move semantics got in the way in such a trivial example.
To convert an rvalue to an lvalue, you can use this lvalue helper function:
template<class T>
T& lvalue_ref(T&& x) { return x; }
And then the call becomes:
scan(lvalue_ref(std::ifstream("myfile")), lvalue_ref(Handler()));
This is safe as the temporaries (the ifstream and Handler) aren't destructed until the end of the full expression. However, note that these are lvalue references to temporaries and as such you must use caution when deciding to use this method. I'm assuming the scan() doesn't hold references/pointers to the arguments after it returns.
For example, do not use it like this:
int& x = lvalue_ref(5);
std::cout << x; // temporary is destructed, therefore Undefined Behavior
Just make sure the lifetime of the returned reference corresponds with the lifetime of the temporary, and you'll be fine.
I was wondering if there is a point in having a rvalue reference variable (not as a function parameter)? I understand the use of rvalue reference when it is used as a function variable as then it is possible to avoid unnecessary allocations etc. But is there a use case rvalue reference variables that are not function parameters? Initially I thought that with such variables we could capture data that was passed in as rvalue for use later, but it seems that if we have a rvalue reference variable, then we can already take its address and therefore it can not be rvalue reference anymore.
I tried the following code and it ( unsurprisingly ) does not compile. So why would I ever want to have a rvalue reference variable?
void test(int&& a) {}
int&& a(22);
test(a);
Thanks!
Your a here is the name of a rvalue reference. a itself is a lvalue (you can take its address) of type rvalue reference (yes, a bit confusing). So whenever you pass your a to test(a), you pass a lvalue. Think about lvalues as any object that has a name, or of which you can latter take its address. And you cannot bind a lvalue to a rvalue reference, hence the compile error
error: cannot bind 'int' lvalue to 'int&&'
You need to pass std::move(a). In this case, you cast away the lvalue-ness of a.
Note that inside your function a is also a lvalue (again, you can perform lvalue operations on it, like taking its address etc). You need a cast (std::move) anytime you want to use it as a rvalue. For example, say you have an object Bar with a member variable Foo member;, and you want to move into it via a member function f:
void Bar::f(Foo&& param)
{
member = std::move(param); // need std::move here to move param into member
}
// invoked it as
bar.f(std::move(some_foo)); // move some_foo into param
If you don't cast param to a rvalue reference with std::move, then you don't move param into member, you just copy it (if param is copy-able, otherwise you get a compile-time error), since param itself is a lvalue. Hope this clarifies it.
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.