Related
I am confused about using std::move() in below code:
If I uncomment line at (2) the output would be: 1 2 3 but if I uncomment line at (1) output would be nothing which means that move constructor of std::vector was called!
Why do we have to make another call to std::move at (1) to make move constructor of std::vector to be called?
What I understood that std::move get the r-value of its parameter so, why we have to get the r-value of r-value at (1)?
I think this line _v = rv; at (2) is more logical and should make std::vector move constructor to be called without std::movebecause rv itself is r-value reference in the first place.
template <class T>
class A
{
public:
void set(std::vector<T> & lv)
{
}
void set(std::vector<T> && rv)
{
//_v = std::move(rv); (1)
//_v = rv; (2)
}
private:
std::vector<T> _v;
};
int main()
{
std::vector<int> vec{1,2,3};
A<int> a;
a.set(std::move(vec));
for(auto &item : vec)
cout << item << " ";
cout << endl;
return 0;
}
Every named object is a Lvalue:
the name of a variable, a function, a template parameter object (since
C++20), or a data member, regardless of type, such as std::cin or
std::endl. Even if the variable's type is rvalue reference, the
expression consisting of its name is an lvalue expression;
vector has two overloads of assignment operator, one for Lvalue reference and another for Rvalue reference.
vector::operator=(const vector&) // copy assignment operator
vector::operator=(vector&&) // move assignment operator
Overload which takes Lvalue reference is called when Lvalue is passed as argument for operator=.
Details here
when a function has both rvalue reference and lvalue reference
overloads, the rvalue reference overload binds to rvalues (including
both prvalues and xvalues), while the lvalue reference overload binds
to lvalues
By std::move(rv); you cast rv - Lvalue to Rvalue reference, and operator= which takes Rvalue reference is called. Otherwise, Lvalue binds to Lvalue reference and vector is copied instead of being moved.
I've been looking into some of the new features of C++11 and one I've noticed is the double ampersand in declaring variables, like T&& var.
For a start, what is this beast called? I wish Google would allow us to search for punctuation like this.
What exactly does it mean?
At first glance, it appears to be a double reference (like the C-style double pointers T** var), but I'm having a hard time thinking of a use case for that.
It declares an rvalue reference (standards proposal doc).
Here's an introduction to rvalue references.
Here's a fantastic in-depth look at rvalue references by one of Microsoft's standard library developers.
CAUTION: the linked article on MSDN ("Rvalue References: C++0x Features in VC10, Part 2") is a very clear introduction to Rvalue references, but makes statements about Rvalue references that were once true in the draft C++11 standard, but are not true for the final one! Specifically, it says at various points that rvalue references can bind to lvalues, which was once true, but was changed.(e.g. int x; int &&rrx = x; no longer compiles in GCC) – drewbarbs Jul 13 '14 at 16:12
The biggest difference between a C++03 reference (now called an lvalue reference in C++11) is that it can bind to an rvalue like a temporary without having to be const. Thus, this syntax is now legal:
T&& r = T();
rvalue references primarily provide for the following:
Move semantics. A move constructor and move assignment operator can now be defined that takes an rvalue reference instead of the usual const-lvalue reference. A move functions like a copy, except it is not obliged to keep the source unchanged; in fact, it usually modifies the source such that it no longer owns the moved resources. This is great for eliminating extraneous copies, especially in standard library implementations.
For example, a copy constructor might look like this:
foo(foo const& other)
{
this->length = other.length;
this->ptr = new int[other.length];
copy(other.ptr, other.ptr + other.length, this->ptr);
}
If this constructor were passed a temporary, the copy would be unnecessary because we know the temporary will just be destroyed; why not make use of the resources the temporary already allocated? In C++03, there's no way to prevent the copy as we cannot determine whether we were passed a temporary. In C++11, we can overload a move constructor:
foo(foo&& other)
{
this->length = other.length;
this->ptr = other.ptr;
other.length = 0;
other.ptr = nullptr;
}
Notice the big difference here: the move constructor actually modifies its argument. This would effectively "move" the temporary into the object being constructed, thereby eliminating the unnecessary copy.
The move constructor would be used for temporaries and for non-const lvalue references that are explicitly converted to rvalue references using the std::move function (it just performs the conversion). The following code both invoke the move constructor for f1 and f2:
foo f1((foo())); // Move a temporary into f1; temporary becomes "empty"
foo f2 = std::move(f1); // Move f1 into f2; f1 is now "empty"
Perfect forwarding. rvalue references allow us to properly forward arguments for templated functions. Take for example this factory function:
template <typename T, typename A1>
std::unique_ptr<T> factory(A1& a1)
{
return std::unique_ptr<T>(new T(a1));
}
If we called factory<foo>(5), the argument will be deduced to be int&, which will not bind to a literal 5, even if foo's constructor takes an int. Well, we could instead use A1 const&, but what if foo takes the constructor argument by non-const reference? To make a truly generic factory function, we would have to overload factory on A1& and on A1 const&. That might be fine if factory takes 1 parameter type, but each additional parameter type would multiply the necessary overload set by 2. That's very quickly unmaintainable.
rvalue references fix this problem by allowing the standard library to define a std::forward function that can properly forward lvalue/rvalue references. For more information about how std::forward works, see this excellent answer.
This enables us to define the factory function like this:
template <typename T, typename A1>
std::unique_ptr<T> factory(A1&& a1)
{
return std::unique_ptr<T>(new T(std::forward<A1>(a1)));
}
Now the argument's rvalue/lvalue-ness is preserved when passed to T's constructor. That means that if factory is called with an rvalue, T's constructor is called with an rvalue. If factory is called with an lvalue, T's constructor is called with an lvalue. The improved factory function works because of one special rule:
When the function parameter type is of
the form T&& where T is a template
parameter, and the function argument
is an lvalue of type A, the type A& is
used for template argument deduction.
Thus, we can use factory like so:
auto p1 = factory<foo>(foo()); // calls foo(foo&&)
auto p2 = factory<foo>(*p1); // calls foo(foo const&)
Important rvalue reference properties:
For overload resolution, lvalues prefer binding to lvalue references and rvalues prefer binding to rvalue references. Hence why temporaries prefer invoking a move constructor / move assignment operator over a copy constructor / assignment operator.
rvalue references will implicitly bind to rvalues and to temporaries that are the result of an implicit conversion. i.e. float f = 0f; int&& i = f; is well formed because float is implicitly convertible to int; the reference would be to a temporary that is the result of the conversion.
Named rvalue references are lvalues. Unnamed rvalue references are rvalues. This is important to understand why the std::move call is necessary in: foo&& r = foo(); foo f = std::move(r);
It denotes an rvalue reference. Rvalue references will only bind to temporary objects, unless explicitly generated otherwise. They are used to make objects much more efficient under certain circumstances, and to provide a facility known as perfect forwarding, which greatly simplifies template code.
In C++03, you can't distinguish between a copy of a non-mutable lvalue and an rvalue.
std::string s;
std::string another(s); // calls std::string(const std::string&);
std::string more(std::string(s)); // calls std::string(const std::string&);
In C++0x, this is not the case.
std::string s;
std::string another(s); // calls std::string(const std::string&);
std::string more(std::string(s)); // calls std::string(std::string&&);
Consider the implementation behind these constructors. In the first case, the string has to perform a copy to retain value semantics, which involves a new heap allocation. However, in the second case, we know in advance that the object which was passed in to our constructor is immediately due for destruction, and it doesn't have to remain untouched. We can effectively just swap the internal pointers and not perform any copying at all in this scenario, which is substantially more efficient. Move semantics benefit any class which has expensive or prohibited copying of internally referenced resources. Consider the case of std::unique_ptr- now that our class can distinguish between temporaries and non-temporaries, we can make the move semantics work correctly so that the unique_ptr cannot be copied but can be moved, which means that std::unique_ptr can be legally stored in Standard containers, sorted, etc, whereas C++03's std::auto_ptr cannot.
Now we consider the other use of rvalue references- perfect forwarding. Consider the question of binding a reference to a reference.
std::string s;
std::string& ref = s;
(std::string&)& anotherref = ref; // usually expressed via template
Can't recall what C++03 says about this, but in C++0x, the resultant type when dealing with rvalue references is critical. An rvalue reference to a type T, where T is a reference type, becomes a reference of type T.
(std::string&)&& ref // ref is std::string&
(const std::string&)&& ref // ref is const std::string&
(std::string&&)&& ref // ref is std::string&&
(const std::string&&)&& ref // ref is const std::string&&
Consider the simplest template function- min and max. In C++03 you have to overload for all four combinations of const and non-const manually. In C++0x it's just one overload. Combined with variadic templates, this enables perfect forwarding.
template<typename A, typename B> auto min(A&& aref, B&& bref) {
// for example, if you pass a const std::string& as first argument,
// then A becomes const std::string& and by extension, aref becomes
// const std::string&, completely maintaining it's type information.
if (std::forward<A>(aref) < std::forward<B>(bref))
return std::forward<A>(aref);
else
return std::forward<B>(bref);
}
I left off the return type deduction, because I can't recall how it's done offhand, but that min can accept any combination of lvalues, rvalues, const lvalues.
The term for T&& when used with type deduction (such as for perfect forwarding) is known colloquially as a forwarding reference. The term "universal reference" was coined by Scott Meyers in this article, but was later changed.
That is because it may be either r-value or l-value.
Examples are:
// template
template<class T> foo(T&& t) { ... }
// auto
auto&& t = ...;
// typedef
typedef ... T;
T&& t = ...;
// decltype
decltype(...)&& t = ...;
More discussion can be found in the answer for: Syntax for universal references
An rvalue reference is a type that behaves much like the ordinary reference X&, with several exceptions. The most important one is that when it comes to function overload resolution, lvalues prefer old-style lvalue references, whereas rvalues prefer the new rvalue references:
void foo(X& x); // lvalue reference overload
void foo(X&& x); // rvalue reference overload
X x;
X foobar();
foo(x); // argument is lvalue: calls foo(X&)
foo(foobar()); // argument is rvalue: calls foo(X&&)
So what is an rvalue? Anything that is not an lvalue. An lvalue being
an expression that refers to a memory location and allows us to take the address of that memory location via the & operator.
It is almost easier to understand first what rvalues accomplish with an example:
#include <cstring>
class Sample {
int *ptr; // large block of memory
int size;
public:
Sample(int sz=0) : ptr{sz != 0 ? new int[sz] : nullptr}, size{sz}
{
if (ptr != nullptr) memset(ptr, 0, sz);
}
// copy constructor that takes lvalue
Sample(const Sample& s) : ptr{s.size != 0 ? new int[s.size] :\
nullptr}, size{s.size}
{
if (ptr != nullptr) memcpy(ptr, s.ptr, s.size);
std::cout << "copy constructor called on lvalue\n";
}
// move constructor that take rvalue
Sample(Sample&& s)
{ // steal s's resources
ptr = s.ptr;
size = s.size;
s.ptr = nullptr; // destructive write
s.size = 0;
cout << "Move constructor called on rvalue." << std::endl;
}
// normal copy assignment operator taking lvalue
Sample& operator=(const Sample& s)
{
if(this != &s) {
delete [] ptr; // free current pointer
size = s.size;
if (size != 0) {
ptr = new int[s.size];
memcpy(ptr, s.ptr, s.size);
} else
ptr = nullptr;
}
cout << "Copy Assignment called on lvalue." << std::endl;
return *this;
}
// overloaded move assignment operator taking rvalue
Sample& operator=(Sample&& lhs)
{
if(this != &s) {
delete [] ptr; //don't let ptr be orphaned
ptr = lhs.ptr; //but now "steal" lhs, don't clone it.
size = lhs.size;
lhs.ptr = nullptr; // lhs's new "stolen" state
lhs.size = 0;
}
cout << "Move Assignment called on rvalue" << std::endl;
return *this;
}
//...snip
};
The constructor and assignment operators have been overloaded with versions that take rvalue references. Rvalue references allow a function to branch at compile time (via overload resolution) on the condition "Am I being called on an lvalue or an rvalue?". This allowed us to create more efficient constructor and assignment operators above that move resources rather copy them.
The compiler automatically branches at compile time (depending on the whether it is being invoked for an lvalue or an rvalue) choosing whether the move constructor or move assignment operator should be called.
Summing up: rvalue references allow move semantics (and perfect forwarding, discussed in the article link below).
One practical easy-to-understand example is the class template std::unique_ptr. Since a unique_ptr maintains exclusive ownership of its underlying raw pointer, unique_ptr's can't be copied. That would violate their invariant of exclusive ownership. So they do not have copy constructors. But they do have move constructors:
template<class T> class unique_ptr {
//...snip
unique_ptr(unique_ptr&& __u) noexcept; // move constructor
};
std::unique_ptr<int[] pt1{new int[10]};
std::unique_ptr<int[]> ptr2{ptr1};// compile error: no copy ctor.
// So we must first cast ptr1 to an rvalue
std::unique_ptr<int[]> ptr2{std::move(ptr1)};
std::unique_ptr<int[]> TakeOwnershipAndAlter(std::unique_ptr<int[]> param,\
int size)
{
for (auto i = 0; i < size; ++i) {
param[i] += 10;
}
return param; // implicitly calls unique_ptr(unique_ptr&&)
}
// Now use function
unique_ptr<int[]> ptr{new int[10]};
// first cast ptr from lvalue to rvalue
unique_ptr<int[]> new_owner = TakeOwnershipAndAlter(\
static_cast<unique_ptr<int[]>&&>(ptr), 10);
cout << "output:\n";
for(auto i = 0; i< 10; ++i) {
cout << new_owner[i] << ", ";
}
output:
10, 10, 10, 10, 10, 10, 10, 10, 10, 10,
static_cast<unique_ptr<int[]>&&>(ptr) is usually done using std::move
// first cast ptr from lvalue to rvalue
unique_ptr<int[]> new_owner = TakeOwnershipAndAlter(std::move(ptr),0);
An excellent article explaining all this and more (like how rvalues allow perfect forwarding and what that means) with lots of good examples is Thomas Becker's C++ Rvalue References Explained. This post relied heavily on his article.
A shorter introduction is A Brief Introduction to Rvalue References by Stroutrup, et. al
Following on from a comment I made on this:
passing std::vector to constructor and move semantics
Is the std::move necessary in the following code, to ensure that the returned value is a xvalue?
std::vector<string> buildVector()
{
std::vector<string> local;
// .... build a vector
return std::move(local);
}
It is my understanding that this is required. I have often seen this used when returning a std::unique_ptr from a function, however GManNickG made the following comment:
It is my understanding that in a return statement all local variables are automatically xvalues (expiring values) and will be moved, but I'm unsure if that only applies to the returned object itself. So OP should go ahead and put that in there until I'm more confident it shouldn't have to be. :)
Can anyone clarify if the std::move is necessary?
Is the behaviour compiler dependent?
You're guaranteed that local will be returned as an rvalue in this situation. Usually compilers would perform return-value optimization though before this even becomes an issue, and you probably wouldn't see any actual move at all, since the local object would be constructed directly at the call site.
A relevant Note in 6.6.3 ["The return statement"] (2):
A copy or move operation associated with a return statement may be elided or considered as an rvalue for the purpose of overload resolution in selecting a constructor (12.8).
To clarify, this is to say that the returned object can be move-constructed from the local object (even though in practice RVO will skip this step entirely). The normative part of the standard is 12.8 ["Copying and moving class objects"] (31, 32), on copy elision and rvalues (thanks #Mankarse!).
Here's a silly example:
#include <utility>
struct Foo
{
Foo() = default;
Foo(Foo const &) = delete;
Foo(Foo &&) = default;
};
Foo f(Foo & x)
{
Foo y;
// return x; // error: use of deleted function ‘Foo::Foo(const Foo&)’
return std::move(x); // OK
return std::move(y); // OK
return y; // OK (!!)
}
Contrast this with returning an actual rvalue reference:
Foo && g()
{
Foo y;
// return y; // error: cannot bind ‘Foo’ lvalue to ‘Foo&&’
return std::move(y); // OK type-wise (but undefined behaviour, thanks #GMNG)
}
Altough both, return std::move(local) and return local, do work in sense of that they do compile, their behavior is different. And probably only the latter one was intended.
If you write a function which returns a std::vector<string>, you have to return a std::vector<string> and exactly it. std::move(local) has the typestd::vector<string>&& which is not a std::vector<string> so it has to be converted to it using the move constructor.
The standard says in 6.6.3.2:
The value of the expression is implicitly
converted to the return type of the function in which it appears.
That means, return std::move(local) is equalvalent to
std::vector<std::string> converted(std::move(local); // move constructor
return converted; // not yet a copy constructor call (which will be elided anyway)
whereas return local only is
return local; // not yet a copy constructor call (which will be elided anyway)
This spares you one operation.
To give you a short example of what that means:
struct test {
test() { std::cout << " construct\n"; }
test(const test&) { std::cout << " copy\n"; }
test(test&&) { std::cout << " move\n"; }
};
test f1() { test t; return t; }
test f2() { test t; return std::move(t); }
int main()
{
std::cout << "f1():\n"; test t1 = f1();
std::cout << "f2():\n"; test t2 = f2();
}
This will output
f1():
construct
f2():
construct
move
I think the answer is no. Though officially only a note, §5/6 summarizes what expressions are/aren't xvalues:
An expression is an xvalue if it is:
the result of calling a function, whether implicitly or explicitly, whose return type is an rvalue reference to object type,
a cast to an rvalue reference to object type,
a class member access expression designating a non-static data member of non-reference type in which the object expression is an xvalue, or
a .* pointer-to-member expression in which the first operand is an xvalue and the second operand is a pointer to data member.
In general, the effect of this rule is that named rvalue references are treated as lvalues and unnamed rvalue references to objects are treated as xvalues; rvalue references to functions are treated as lvalues whether named or not.
The first bullet point seems to apply here. Since the function in question returns a value rather than an rvalue reference, the result won't be an xvalue.
I've been looking into some of the new features of C++11 and one I've noticed is the double ampersand in declaring variables, like T&& var.
For a start, what is this beast called? I wish Google would allow us to search for punctuation like this.
What exactly does it mean?
At first glance, it appears to be a double reference (like the C-style double pointers T** var), but I'm having a hard time thinking of a use case for that.
It declares an rvalue reference (standards proposal doc).
Here's an introduction to rvalue references.
Here's a fantastic in-depth look at rvalue references by one of Microsoft's standard library developers.
CAUTION: the linked article on MSDN ("Rvalue References: C++0x Features in VC10, Part 2") is a very clear introduction to Rvalue references, but makes statements about Rvalue references that were once true in the draft C++11 standard, but are not true for the final one! Specifically, it says at various points that rvalue references can bind to lvalues, which was once true, but was changed.(e.g. int x; int &&rrx = x; no longer compiles in GCC) – drewbarbs Jul 13 '14 at 16:12
The biggest difference between a C++03 reference (now called an lvalue reference in C++11) is that it can bind to an rvalue like a temporary without having to be const. Thus, this syntax is now legal:
T&& r = T();
rvalue references primarily provide for the following:
Move semantics. A move constructor and move assignment operator can now be defined that takes an rvalue reference instead of the usual const-lvalue reference. A move functions like a copy, except it is not obliged to keep the source unchanged; in fact, it usually modifies the source such that it no longer owns the moved resources. This is great for eliminating extraneous copies, especially in standard library implementations.
For example, a copy constructor might look like this:
foo(foo const& other)
{
this->length = other.length;
this->ptr = new int[other.length];
copy(other.ptr, other.ptr + other.length, this->ptr);
}
If this constructor were passed a temporary, the copy would be unnecessary because we know the temporary will just be destroyed; why not make use of the resources the temporary already allocated? In C++03, there's no way to prevent the copy as we cannot determine whether we were passed a temporary. In C++11, we can overload a move constructor:
foo(foo&& other)
{
this->length = other.length;
this->ptr = other.ptr;
other.length = 0;
other.ptr = nullptr;
}
Notice the big difference here: the move constructor actually modifies its argument. This would effectively "move" the temporary into the object being constructed, thereby eliminating the unnecessary copy.
The move constructor would be used for temporaries and for non-const lvalue references that are explicitly converted to rvalue references using the std::move function (it just performs the conversion). The following code both invoke the move constructor for f1 and f2:
foo f1((foo())); // Move a temporary into f1; temporary becomes "empty"
foo f2 = std::move(f1); // Move f1 into f2; f1 is now "empty"
Perfect forwarding. rvalue references allow us to properly forward arguments for templated functions. Take for example this factory function:
template <typename T, typename A1>
std::unique_ptr<T> factory(A1& a1)
{
return std::unique_ptr<T>(new T(a1));
}
If we called factory<foo>(5), the argument will be deduced to be int&, which will not bind to a literal 5, even if foo's constructor takes an int. Well, we could instead use A1 const&, but what if foo takes the constructor argument by non-const reference? To make a truly generic factory function, we would have to overload factory on A1& and on A1 const&. That might be fine if factory takes 1 parameter type, but each additional parameter type would multiply the necessary overload set by 2. That's very quickly unmaintainable.
rvalue references fix this problem by allowing the standard library to define a std::forward function that can properly forward lvalue/rvalue references. For more information about how std::forward works, see this excellent answer.
This enables us to define the factory function like this:
template <typename T, typename A1>
std::unique_ptr<T> factory(A1&& a1)
{
return std::unique_ptr<T>(new T(std::forward<A1>(a1)));
}
Now the argument's rvalue/lvalue-ness is preserved when passed to T's constructor. That means that if factory is called with an rvalue, T's constructor is called with an rvalue. If factory is called with an lvalue, T's constructor is called with an lvalue. The improved factory function works because of one special rule:
When the function parameter type is of
the form T&& where T is a template
parameter, and the function argument
is an lvalue of type A, the type A& is
used for template argument deduction.
Thus, we can use factory like so:
auto p1 = factory<foo>(foo()); // calls foo(foo&&)
auto p2 = factory<foo>(*p1); // calls foo(foo const&)
Important rvalue reference properties:
For overload resolution, lvalues prefer binding to lvalue references and rvalues prefer binding to rvalue references. Hence why temporaries prefer invoking a move constructor / move assignment operator over a copy constructor / assignment operator.
rvalue references will implicitly bind to rvalues and to temporaries that are the result of an implicit conversion. i.e. float f = 0f; int&& i = f; is well formed because float is implicitly convertible to int; the reference would be to a temporary that is the result of the conversion.
Named rvalue references are lvalues. Unnamed rvalue references are rvalues. This is important to understand why the std::move call is necessary in: foo&& r = foo(); foo f = std::move(r);
It denotes an rvalue reference. Rvalue references will only bind to temporary objects, unless explicitly generated otherwise. They are used to make objects much more efficient under certain circumstances, and to provide a facility known as perfect forwarding, which greatly simplifies template code.
In C++03, you can't distinguish between a copy of a non-mutable lvalue and an rvalue.
std::string s;
std::string another(s); // calls std::string(const std::string&);
std::string more(std::string(s)); // calls std::string(const std::string&);
In C++0x, this is not the case.
std::string s;
std::string another(s); // calls std::string(const std::string&);
std::string more(std::string(s)); // calls std::string(std::string&&);
Consider the implementation behind these constructors. In the first case, the string has to perform a copy to retain value semantics, which involves a new heap allocation. However, in the second case, we know in advance that the object which was passed in to our constructor is immediately due for destruction, and it doesn't have to remain untouched. We can effectively just swap the internal pointers and not perform any copying at all in this scenario, which is substantially more efficient. Move semantics benefit any class which has expensive or prohibited copying of internally referenced resources. Consider the case of std::unique_ptr- now that our class can distinguish between temporaries and non-temporaries, we can make the move semantics work correctly so that the unique_ptr cannot be copied but can be moved, which means that std::unique_ptr can be legally stored in Standard containers, sorted, etc, whereas C++03's std::auto_ptr cannot.
Now we consider the other use of rvalue references- perfect forwarding. Consider the question of binding a reference to a reference.
std::string s;
std::string& ref = s;
(std::string&)& anotherref = ref; // usually expressed via template
Can't recall what C++03 says about this, but in C++0x, the resultant type when dealing with rvalue references is critical. An rvalue reference to a type T, where T is a reference type, becomes a reference of type T.
(std::string&)&& ref // ref is std::string&
(const std::string&)&& ref // ref is const std::string&
(std::string&&)&& ref // ref is std::string&&
(const std::string&&)&& ref // ref is const std::string&&
Consider the simplest template function- min and max. In C++03 you have to overload for all four combinations of const and non-const manually. In C++0x it's just one overload. Combined with variadic templates, this enables perfect forwarding.
template<typename A, typename B> auto min(A&& aref, B&& bref) {
// for example, if you pass a const std::string& as first argument,
// then A becomes const std::string& and by extension, aref becomes
// const std::string&, completely maintaining it's type information.
if (std::forward<A>(aref) < std::forward<B>(bref))
return std::forward<A>(aref);
else
return std::forward<B>(bref);
}
I left off the return type deduction, because I can't recall how it's done offhand, but that min can accept any combination of lvalues, rvalues, const lvalues.
The term for T&& when used with type deduction (such as for perfect forwarding) is known colloquially as a forwarding reference. The term "universal reference" was coined by Scott Meyers in this article, but was later changed.
That is because it may be either r-value or l-value.
Examples are:
// template
template<class T> foo(T&& t) { ... }
// auto
auto&& t = ...;
// typedef
typedef ... T;
T&& t = ...;
// decltype
decltype(...)&& t = ...;
More discussion can be found in the answer for: Syntax for universal references
An rvalue reference is a type that behaves much like the ordinary reference X&, with several exceptions. The most important one is that when it comes to function overload resolution, lvalues prefer old-style lvalue references, whereas rvalues prefer the new rvalue references:
void foo(X& x); // lvalue reference overload
void foo(X&& x); // rvalue reference overload
X x;
X foobar();
foo(x); // argument is lvalue: calls foo(X&)
foo(foobar()); // argument is rvalue: calls foo(X&&)
So what is an rvalue? Anything that is not an lvalue. An lvalue being
an expression that refers to a memory location and allows us to take the address of that memory location via the & operator.
It is almost easier to understand first what rvalues accomplish with an example:
#include <cstring>
class Sample {
int *ptr; // large block of memory
int size;
public:
Sample(int sz=0) : ptr{sz != 0 ? new int[sz] : nullptr}, size{sz}
{
if (ptr != nullptr) memset(ptr, 0, sz);
}
// copy constructor that takes lvalue
Sample(const Sample& s) : ptr{s.size != 0 ? new int[s.size] :\
nullptr}, size{s.size}
{
if (ptr != nullptr) memcpy(ptr, s.ptr, s.size);
std::cout << "copy constructor called on lvalue\n";
}
// move constructor that take rvalue
Sample(Sample&& s)
{ // steal s's resources
ptr = s.ptr;
size = s.size;
s.ptr = nullptr; // destructive write
s.size = 0;
cout << "Move constructor called on rvalue." << std::endl;
}
// normal copy assignment operator taking lvalue
Sample& operator=(const Sample& s)
{
if(this != &s) {
delete [] ptr; // free current pointer
size = s.size;
if (size != 0) {
ptr = new int[s.size];
memcpy(ptr, s.ptr, s.size);
} else
ptr = nullptr;
}
cout << "Copy Assignment called on lvalue." << std::endl;
return *this;
}
// overloaded move assignment operator taking rvalue
Sample& operator=(Sample&& lhs)
{
if(this != &s) {
delete [] ptr; //don't let ptr be orphaned
ptr = lhs.ptr; //but now "steal" lhs, don't clone it.
size = lhs.size;
lhs.ptr = nullptr; // lhs's new "stolen" state
lhs.size = 0;
}
cout << "Move Assignment called on rvalue" << std::endl;
return *this;
}
//...snip
};
The constructor and assignment operators have been overloaded with versions that take rvalue references. Rvalue references allow a function to branch at compile time (via overload resolution) on the condition "Am I being called on an lvalue or an rvalue?". This allowed us to create more efficient constructor and assignment operators above that move resources rather copy them.
The compiler automatically branches at compile time (depending on the whether it is being invoked for an lvalue or an rvalue) choosing whether the move constructor or move assignment operator should be called.
Summing up: rvalue references allow move semantics (and perfect forwarding, discussed in the article link below).
One practical easy-to-understand example is the class template std::unique_ptr. Since a unique_ptr maintains exclusive ownership of its underlying raw pointer, unique_ptr's can't be copied. That would violate their invariant of exclusive ownership. So they do not have copy constructors. But they do have move constructors:
template<class T> class unique_ptr {
//...snip
unique_ptr(unique_ptr&& __u) noexcept; // move constructor
};
std::unique_ptr<int[] pt1{new int[10]};
std::unique_ptr<int[]> ptr2{ptr1};// compile error: no copy ctor.
// So we must first cast ptr1 to an rvalue
std::unique_ptr<int[]> ptr2{std::move(ptr1)};
std::unique_ptr<int[]> TakeOwnershipAndAlter(std::unique_ptr<int[]> param,\
int size)
{
for (auto i = 0; i < size; ++i) {
param[i] += 10;
}
return param; // implicitly calls unique_ptr(unique_ptr&&)
}
// Now use function
unique_ptr<int[]> ptr{new int[10]};
// first cast ptr from lvalue to rvalue
unique_ptr<int[]> new_owner = TakeOwnershipAndAlter(\
static_cast<unique_ptr<int[]>&&>(ptr), 10);
cout << "output:\n";
for(auto i = 0; i< 10; ++i) {
cout << new_owner[i] << ", ";
}
output:
10, 10, 10, 10, 10, 10, 10, 10, 10, 10,
static_cast<unique_ptr<int[]>&&>(ptr) is usually done using std::move
// first cast ptr from lvalue to rvalue
unique_ptr<int[]> new_owner = TakeOwnershipAndAlter(std::move(ptr),0);
An excellent article explaining all this and more (like how rvalues allow perfect forwarding and what that means) with lots of good examples is Thomas Becker's C++ Rvalue References Explained. This post relied heavily on his article.
A shorter introduction is A Brief Introduction to Rvalue References by Stroutrup, et. al
I've been looking into some of the new features of C++11 and one I've noticed is the double ampersand in declaring variables, like T&& var.
For a start, what is this beast called? I wish Google would allow us to search for punctuation like this.
What exactly does it mean?
At first glance, it appears to be a double reference (like the C-style double pointers T** var), but I'm having a hard time thinking of a use case for that.
It declares an rvalue reference (standards proposal doc).
Here's an introduction to rvalue references.
Here's a fantastic in-depth look at rvalue references by one of Microsoft's standard library developers.
CAUTION: the linked article on MSDN ("Rvalue References: C++0x Features in VC10, Part 2") is a very clear introduction to Rvalue references, but makes statements about Rvalue references that were once true in the draft C++11 standard, but are not true for the final one! Specifically, it says at various points that rvalue references can bind to lvalues, which was once true, but was changed.(e.g. int x; int &&rrx = x; no longer compiles in GCC) – drewbarbs Jul 13 '14 at 16:12
The biggest difference between a C++03 reference (now called an lvalue reference in C++11) is that it can bind to an rvalue like a temporary without having to be const. Thus, this syntax is now legal:
T&& r = T();
rvalue references primarily provide for the following:
Move semantics. A move constructor and move assignment operator can now be defined that takes an rvalue reference instead of the usual const-lvalue reference. A move functions like a copy, except it is not obliged to keep the source unchanged; in fact, it usually modifies the source such that it no longer owns the moved resources. This is great for eliminating extraneous copies, especially in standard library implementations.
For example, a copy constructor might look like this:
foo(foo const& other)
{
this->length = other.length;
this->ptr = new int[other.length];
copy(other.ptr, other.ptr + other.length, this->ptr);
}
If this constructor were passed a temporary, the copy would be unnecessary because we know the temporary will just be destroyed; why not make use of the resources the temporary already allocated? In C++03, there's no way to prevent the copy as we cannot determine whether we were passed a temporary. In C++11, we can overload a move constructor:
foo(foo&& other)
{
this->length = other.length;
this->ptr = other.ptr;
other.length = 0;
other.ptr = nullptr;
}
Notice the big difference here: the move constructor actually modifies its argument. This would effectively "move" the temporary into the object being constructed, thereby eliminating the unnecessary copy.
The move constructor would be used for temporaries and for non-const lvalue references that are explicitly converted to rvalue references using the std::move function (it just performs the conversion). The following code both invoke the move constructor for f1 and f2:
foo f1((foo())); // Move a temporary into f1; temporary becomes "empty"
foo f2 = std::move(f1); // Move f1 into f2; f1 is now "empty"
Perfect forwarding. rvalue references allow us to properly forward arguments for templated functions. Take for example this factory function:
template <typename T, typename A1>
std::unique_ptr<T> factory(A1& a1)
{
return std::unique_ptr<T>(new T(a1));
}
If we called factory<foo>(5), the argument will be deduced to be int&, which will not bind to a literal 5, even if foo's constructor takes an int. Well, we could instead use A1 const&, but what if foo takes the constructor argument by non-const reference? To make a truly generic factory function, we would have to overload factory on A1& and on A1 const&. That might be fine if factory takes 1 parameter type, but each additional parameter type would multiply the necessary overload set by 2. That's very quickly unmaintainable.
rvalue references fix this problem by allowing the standard library to define a std::forward function that can properly forward lvalue/rvalue references. For more information about how std::forward works, see this excellent answer.
This enables us to define the factory function like this:
template <typename T, typename A1>
std::unique_ptr<T> factory(A1&& a1)
{
return std::unique_ptr<T>(new T(std::forward<A1>(a1)));
}
Now the argument's rvalue/lvalue-ness is preserved when passed to T's constructor. That means that if factory is called with an rvalue, T's constructor is called with an rvalue. If factory is called with an lvalue, T's constructor is called with an lvalue. The improved factory function works because of one special rule:
When the function parameter type is of
the form T&& where T is a template
parameter, and the function argument
is an lvalue of type A, the type A& is
used for template argument deduction.
Thus, we can use factory like so:
auto p1 = factory<foo>(foo()); // calls foo(foo&&)
auto p2 = factory<foo>(*p1); // calls foo(foo const&)
Important rvalue reference properties:
For overload resolution, lvalues prefer binding to lvalue references and rvalues prefer binding to rvalue references. Hence why temporaries prefer invoking a move constructor / move assignment operator over a copy constructor / assignment operator.
rvalue references will implicitly bind to rvalues and to temporaries that are the result of an implicit conversion. i.e. float f = 0f; int&& i = f; is well formed because float is implicitly convertible to int; the reference would be to a temporary that is the result of the conversion.
Named rvalue references are lvalues. Unnamed rvalue references are rvalues. This is important to understand why the std::move call is necessary in: foo&& r = foo(); foo f = std::move(r);
It denotes an rvalue reference. Rvalue references will only bind to temporary objects, unless explicitly generated otherwise. They are used to make objects much more efficient under certain circumstances, and to provide a facility known as perfect forwarding, which greatly simplifies template code.
In C++03, you can't distinguish between a copy of a non-mutable lvalue and an rvalue.
std::string s;
std::string another(s); // calls std::string(const std::string&);
std::string more(std::string(s)); // calls std::string(const std::string&);
In C++0x, this is not the case.
std::string s;
std::string another(s); // calls std::string(const std::string&);
std::string more(std::string(s)); // calls std::string(std::string&&);
Consider the implementation behind these constructors. In the first case, the string has to perform a copy to retain value semantics, which involves a new heap allocation. However, in the second case, we know in advance that the object which was passed in to our constructor is immediately due for destruction, and it doesn't have to remain untouched. We can effectively just swap the internal pointers and not perform any copying at all in this scenario, which is substantially more efficient. Move semantics benefit any class which has expensive or prohibited copying of internally referenced resources. Consider the case of std::unique_ptr- now that our class can distinguish between temporaries and non-temporaries, we can make the move semantics work correctly so that the unique_ptr cannot be copied but can be moved, which means that std::unique_ptr can be legally stored in Standard containers, sorted, etc, whereas C++03's std::auto_ptr cannot.
Now we consider the other use of rvalue references- perfect forwarding. Consider the question of binding a reference to a reference.
std::string s;
std::string& ref = s;
(std::string&)& anotherref = ref; // usually expressed via template
Can't recall what C++03 says about this, but in C++0x, the resultant type when dealing with rvalue references is critical. An rvalue reference to a type T, where T is a reference type, becomes a reference of type T.
(std::string&)&& ref // ref is std::string&
(const std::string&)&& ref // ref is const std::string&
(std::string&&)&& ref // ref is std::string&&
(const std::string&&)&& ref // ref is const std::string&&
Consider the simplest template function- min and max. In C++03 you have to overload for all four combinations of const and non-const manually. In C++0x it's just one overload. Combined with variadic templates, this enables perfect forwarding.
template<typename A, typename B> auto min(A&& aref, B&& bref) {
// for example, if you pass a const std::string& as first argument,
// then A becomes const std::string& and by extension, aref becomes
// const std::string&, completely maintaining it's type information.
if (std::forward<A>(aref) < std::forward<B>(bref))
return std::forward<A>(aref);
else
return std::forward<B>(bref);
}
I left off the return type deduction, because I can't recall how it's done offhand, but that min can accept any combination of lvalues, rvalues, const lvalues.
The term for T&& when used with type deduction (such as for perfect forwarding) is known colloquially as a forwarding reference. The term "universal reference" was coined by Scott Meyers in this article, but was later changed.
That is because it may be either r-value or l-value.
Examples are:
// template
template<class T> foo(T&& t) { ... }
// auto
auto&& t = ...;
// typedef
typedef ... T;
T&& t = ...;
// decltype
decltype(...)&& t = ...;
More discussion can be found in the answer for: Syntax for universal references
An rvalue reference is a type that behaves much like the ordinary reference X&, with several exceptions. The most important one is that when it comes to function overload resolution, lvalues prefer old-style lvalue references, whereas rvalues prefer the new rvalue references:
void foo(X& x); // lvalue reference overload
void foo(X&& x); // rvalue reference overload
X x;
X foobar();
foo(x); // argument is lvalue: calls foo(X&)
foo(foobar()); // argument is rvalue: calls foo(X&&)
So what is an rvalue? Anything that is not an lvalue. An lvalue being
an expression that refers to a memory location and allows us to take the address of that memory location via the & operator.
It is almost easier to understand first what rvalues accomplish with an example:
#include <cstring>
class Sample {
int *ptr; // large block of memory
int size;
public:
Sample(int sz=0) : ptr{sz != 0 ? new int[sz] : nullptr}, size{sz}
{
if (ptr != nullptr) memset(ptr, 0, sz);
}
// copy constructor that takes lvalue
Sample(const Sample& s) : ptr{s.size != 0 ? new int[s.size] :\
nullptr}, size{s.size}
{
if (ptr != nullptr) memcpy(ptr, s.ptr, s.size);
std::cout << "copy constructor called on lvalue\n";
}
// move constructor that take rvalue
Sample(Sample&& s)
{ // steal s's resources
ptr = s.ptr;
size = s.size;
s.ptr = nullptr; // destructive write
s.size = 0;
cout << "Move constructor called on rvalue." << std::endl;
}
// normal copy assignment operator taking lvalue
Sample& operator=(const Sample& s)
{
if(this != &s) {
delete [] ptr; // free current pointer
size = s.size;
if (size != 0) {
ptr = new int[s.size];
memcpy(ptr, s.ptr, s.size);
} else
ptr = nullptr;
}
cout << "Copy Assignment called on lvalue." << std::endl;
return *this;
}
// overloaded move assignment operator taking rvalue
Sample& operator=(Sample&& lhs)
{
if(this != &s) {
delete [] ptr; //don't let ptr be orphaned
ptr = lhs.ptr; //but now "steal" lhs, don't clone it.
size = lhs.size;
lhs.ptr = nullptr; // lhs's new "stolen" state
lhs.size = 0;
}
cout << "Move Assignment called on rvalue" << std::endl;
return *this;
}
//...snip
};
The constructor and assignment operators have been overloaded with versions that take rvalue references. Rvalue references allow a function to branch at compile time (via overload resolution) on the condition "Am I being called on an lvalue or an rvalue?". This allowed us to create more efficient constructor and assignment operators above that move resources rather copy them.
The compiler automatically branches at compile time (depending on the whether it is being invoked for an lvalue or an rvalue) choosing whether the move constructor or move assignment operator should be called.
Summing up: rvalue references allow move semantics (and perfect forwarding, discussed in the article link below).
One practical easy-to-understand example is the class template std::unique_ptr. Since a unique_ptr maintains exclusive ownership of its underlying raw pointer, unique_ptr's can't be copied. That would violate their invariant of exclusive ownership. So they do not have copy constructors. But they do have move constructors:
template<class T> class unique_ptr {
//...snip
unique_ptr(unique_ptr&& __u) noexcept; // move constructor
};
std::unique_ptr<int[] pt1{new int[10]};
std::unique_ptr<int[]> ptr2{ptr1};// compile error: no copy ctor.
// So we must first cast ptr1 to an rvalue
std::unique_ptr<int[]> ptr2{std::move(ptr1)};
std::unique_ptr<int[]> TakeOwnershipAndAlter(std::unique_ptr<int[]> param,\
int size)
{
for (auto i = 0; i < size; ++i) {
param[i] += 10;
}
return param; // implicitly calls unique_ptr(unique_ptr&&)
}
// Now use function
unique_ptr<int[]> ptr{new int[10]};
// first cast ptr from lvalue to rvalue
unique_ptr<int[]> new_owner = TakeOwnershipAndAlter(\
static_cast<unique_ptr<int[]>&&>(ptr), 10);
cout << "output:\n";
for(auto i = 0; i< 10; ++i) {
cout << new_owner[i] << ", ";
}
output:
10, 10, 10, 10, 10, 10, 10, 10, 10, 10,
static_cast<unique_ptr<int[]>&&>(ptr) is usually done using std::move
// first cast ptr from lvalue to rvalue
unique_ptr<int[]> new_owner = TakeOwnershipAndAlter(std::move(ptr),0);
An excellent article explaining all this and more (like how rvalues allow perfect forwarding and what that means) with lots of good examples is Thomas Becker's C++ Rvalue References Explained. This post relied heavily on his article.
A shorter introduction is A Brief Introduction to Rvalue References by Stroutrup, et. al