Rvalue Reference Declarator: &&
Holds a reference to an rvalue expression.
type-id && cast-expression
Remarks
Rvalue references enable you to distinguish an lvalue from an rvalue. Lvalue references and rvalue references are syntactically and semantically similar, but they follow somewhat different rules. For more information about lvalues and rvalues, see Lvalues and Rvalues. For more information about lvalue references, see Lvalue Reference Declarator: &.
The following sections describe how rvalue references support the implementation of move semantics and perfect forwarding.
Move Semantics
Rvalue references support the implementation of move semantics, which can significantly increase the performance of your applications. Move semantics enables you to write code that transfers resources (such as dynamically allocated memory) from one object to another. Move semantics works because it enables resources to be transferred from temporary objects that cannot be referenced elsewhere in the program.
To implement move semantics, you typically provide a move constructor, and optionally a move assignment operator (operator=), to your class. Copy and assignment operations whose sources are rvalues then automatically take advantage of move semantics. Unlike the default copy constructor, the compiler does not provide a default move constructor. For more information about how to write a move constructor and how to use it in your application, see How to: Write a Move Constructor.
You can also overload ordinary functions and operators to take advantage of move semantics. Visual C++ 2010 introduces move semantics into the Standard Template Library (STL). For example, the string class implements operations that perform move semantics. Consider the following example that concatenates several strings and prints the result:
// string_concatenation.cpp
// compile with: /EHsc
#include <iostream>
#include <string>
using namespace std;
int main()
{
string s = string("h") + "e" + "ll" + "o";
cout << s << endl;
}
Before Visual C++ 2010, each call to operator+ allocates and returns a new temporary string object (an rvalue). operator+ cannot append one string to the other because it does not know whether the source strings are lvalues or rvalues. If the source strings are both lvalues, they might be referenced elsewhere in the program and therefore must not be modified. By using rvalue references, operator+ can be modified to take rvalues, which cannot be referenced elsewhere in the program. Therefore, operator+ can now append one string to another. This can significantly reduce the number of dynamic memory allocations that the string class must perform. For more information about the string class, see basic_string Class.
Move semantics also helps when the compiler cannot use Return Value Optimization (RVO) or Named Return Value Optimization (NRVO). In these cases, the compiler calls the move constructor if the type defines it. For more information about Named Return Value Optimization, see Named Return Value Optimization in Visual C++ 2005.
To better understand move semantics, consider the example of inserting an element into a vector object. If the capacity of the vector object is exceeded, the vector object must reallocate memory for its elements and then copy each element to another memory location to make room for the inserted element. When an insertion operation copies an element, it creates a new element, calls the copy constructor to copy the data from the previous element to the new element, and then destroys the previous element. Move semantics enables you to move objects directly without having to perform expensive memory allocation and copy operations.
To take advantage of move semantics in the vector example, you can write a move constructor to move data from one object to another.
For more information about the introduction of move semantics into the STL in Visual C++ 2010, see C++ Standard Library Reference.
Perfect Forwarding
Perfect forwarding reduces the need for overloaded functions and helps avoid the forwarding problem. The forwarding problem can occur when you write a generic function that takes references as its parameters and it passes (or forwards) these parameters to another function. For example, if the generic function takes a parameter of type const T&, then the called function cannot modify the value of that parameter. If the generic function takes a parameter of type T&, then the function cannot be called by using an rvalue (such as a temporary object or integer literal).
Ordinarily, to solve this problem, you must provide overloaded versions of the generic function that take both T& and const T& for each of its parameters. As a result, the number of overloaded functions increases exponentially with the number of parameters. Rvalue references enable you to write one version of a function that accepts arbitrary arguments and forwards them to another function as if the other function had been called directly.
Consider the following example that declares four types, W, X, Y, and Z. The constructor for each type takes a different combination of const and non-const lvalue references as its parameters.
struct W
{
W(int&, int&) {}
};
struct X
{
X(const int&, int&) {}
};
struct Y
{
Y(int&, const int&) {}
};
struct Z
{
Z(const int&, const int&) {}
};
Suppose you want to write a generic function that generates objects. The following example shows one way to write this function:
template <typename T, typename A1, typename A2>
T* factory(A1& a1, A2& a2)
{
return new T(a1, a2);
}
The following example shows a valid call to the factory function:
int a = 4, b = 5;
W* pw = factory<W>(a, b);
However, the following example does not contain a valid call to the factory function because factory takes lvalue references that are modifiable as its parameters, but it is called by using rvalues:
Z* pz = factory<Z>(2, 2);
Ordinarily, to solve this problem, you must create an overloaded version of the factory function for every combination of A& and const A& parameters. Rvalue references enable you to write one version of the factory function, as shown in the following example:
template <typename T, typename A1, typename A2>
T* factory(A1&& a1, A2&& a2)
{
return new T(std::forward<A1>(a1), std::forward<A2>(a2));
}
This example uses rvalue references as the parameters to the factory function. The purpose of the std::forward function is to forward the parameters of the factory function to the constructor of the template class.
The following example shows the main function that uses the revised factory function to create instances of the W, X, Y, and Z classes. The revised factory function forwards its parameters (either lvalues or rvalues) to the appropriate class constructor.
int main()
{
int a = 4, b = 5;
W* pw = factory<W>(a, b);
X* px = factory<X>(2, b);
Y* py = factory<Y>(a, 2);
Z* pz = factory<Z>(2, 2);
delete pw;
delete px;
delete py;
delete pz;
}
Additional Properties of Rvalue References
You can overload a function to take an lvalue reference and an rvalue reference.
By overloading a function to take a const lvalue reference or an rvalue reference, you can write code that distinguishes between non-modifiable objects (lvalues) and modifiable temporary values (rvalues). You can pass an object to a function that takes an rvalue reference unless the object is marked as const. The following example shows the function f, which is overloaded to take an lvalue reference and an rvalue reference. The main function calls f with both lvalues and an rvalue.
// reference-overload.cpp
// Compile with: /EHsc
#include <iostream>
using namespace std;
// A class that contains a memory resource.
class MemoryBlock
{
// TODO: Add resources for the class here.
};
void f(const MemoryBlock&)
{
cout << "In f(const MemoryBlock&). This version cannot modify the parameter." << endl;
}
void f(MemoryBlock&&)
{
cout << "In f(MemoryBlock&&). This version can modify the parameter." << endl;
}
int main()
{
MemoryBlock block;
f(block);
f(MemoryBlock());
}
This example produces the following output:
In f(const MemoryBlock&). This version cannot modify the parameter.
In f(MemoryBlock&&). This version can modify the parameter.
In this example, the first call to f passes a local variable (an lvalue) as its argument. The second call to f passes a temporary object as its argument. Because the temporary object cannot be referenced elsewhere in the program, the call binds to the overloaded version of f that takes an rvalue reference, which is free to modify the object.
The compiler treats a named rvalue reference as an lvalue and an unnamed rvalue reference as an rvalue.
When you write a function that takes an rvalue reference as its parameter, that parameter is treated as an lvalue in the body of the function. The compiler treats a named rvalue reference as an lvalue because a named object can be referenced by several parts of a program; it would be dangerous to allow multiple parts of a program to modify or remove resources from that object. For example, if multiple parts of a program try to transfer resources from the same object, only the first part will successfully transfer the resource.
The following example shows the function g, which is overloaded to take an lvalue reference and an rvalue reference. The function f takes an rvalue reference as its parameter (a named rvalue reference) and returns an rvalue reference (an unnamed rvalue reference). In the call to g from f, overload resolution selects the version of g that takes an lvalue reference because the body of f treats its parameter as an lvalue. In the call to g from main, overload resolution selects the version of g that takes an rvalue reference because f returns an rvalue reference.
// named-reference.cpp
// Compile with: /EHsc
#include <iostream>
using namespace std;
// A class that contains a memory resource.
class MemoryBlock
{
// TODO: Add resources for the class here.
};
void g(const MemoryBlock&)
{
cout << "In g(const MemoryBlock&)." << endl;
}
void g(MemoryBlock&&)
{
cout << "In g(MemoryBlock&&)." << endl;
}
MemoryBlock&& f(MemoryBlock&& block)
{
g(block);
return block;
}
int main()
{
g(f(MemoryBlock()));
}
This example produces the following output:
In g(const MemoryBlock&).
In g(MemoryBlock&&).
In this example, the main function passes an rvalue to f. The body of f treats its named parameter as an lvalue. The call from f to g binds the parameter to an lvalue reference (the first overloaded version of g).
- You can cast an lvalue to an rvalue reference.
The STL std::move function enables you to convert an object to an rvalue reference to that object. Alternatively, you can use the static_cast keyword to cast an lvalue to an rvalue reference, as shown in the following example:
// cast-reference.cpp
// Compile with: /EHsc
#include <iostream>
using namespace std;
// A class that contains a memory resource.
class MemoryBlock
{
// TODO: Add resources for the class here.
};
void g(const MemoryBlock&)
{
cout << "In g(const MemoryBlock&)." << endl;
}
void g(MemoryBlock&&)
{
cout << "In g(MemoryBlock&&)." << endl;
}
int main()
{
MemoryBlock block;
g(block);
g(static_cast<MemoryBlock&&>(block));
}
This example produces the following output:
In g(const MemoryBlock&).
In g(MemoryBlock&&).
Function templates deduce their template argument types and then use reference collapsing rules.
It is common to write a function template that passes (or forwards) its parameters to another function. It is important to understand how template type deduction works for function templates that take rvalue references.
If the function argument is an rvalue, the compiler deduces the argument to be an rvalue reference. For example, if you pass an rvalue reference to an object of type X to a template function that takes type T&& as its parameter, template argument deduction deduces T to be X. Therefore, the parameter has type X&&. If the function argument is an lvalue or const lvalue, the compiler deduces its type to be an lvalue reference or const lvalue reference of that type.
The following example declares one structure template and then specializes it for various reference types. The print_type_and_value function takes an rvalue reference as its parameter and forwards it to the appropriate specialized version of the S::print method. The main function demonstrates the various ways to call the S::print method.
// template-type-deduction.cpp
// Compile with: /EHsc
#include <iostream>
#include <string>
using namespace std;
template<typename T> struct S;
// The following structures specialize S by
// lvalue reference (T&), const lvalue reference (const T&),
// rvalue reference (T&&), and const rvalue reference (const T&&).
// Each structure provides a print method that prints the type of
// the structure and its parameter.
template<typename T> struct S<T&> {
static void print(T& t)
{
cout << "print<T&>: " << t << endl;
}
};
template<typename T> struct S<const T&> {
static void print(const T& t)
{
cout << "print<const T&>: " << t << endl;
}
};
template<typename T> struct S<T&&> {
static void print(T&& t)
{
cout << "print<T&&>: " << t << endl;
}
};
template<typename T> struct S<const T&&> {
static void print(const T&& t)
{
cout << "print<const T&&>: " << t << endl;
}
};
// This function forwards its parameter to a specialized
// version of the S type.
template <typename T> void print_type_and_value(T&& t)
{
S<T&&>::print(std::forward<T>(t));
}
// This function returns the constant string "fourth".
const string fourth() { return string("fourth"); }
int main()
{
// The following call resolves to:
// print_type_and_value<string&>(string& && t)
// Which collapses to:
// print_type_and_value<string&>(string& t)
string s1("first");
print_type_and_value(s1);
// The following call resolves to:
// print_type_and_value<const string&>(const string& && t)
// Which collapses to:
// print_type_and_value<const string&>(const string& t)
const string s2("second");
print_type_and_value(s2);
// The following call resolves to:
// print_type_and_value<string&&>(string&& t)
print_type_and_value(string("third"));
// The following call resolves to:
// print_type_and_value<const string&&>(const string&& t)
print_type_and_value(fourth());
}
This example produces the following output:
print<T&>: first
print<const T&>: second
print<T&&>: third
print<const T&&>: fourth
To resolve each call to the print_type_and_value function, the compiler first performs template argument deduction. The compiler then applies reference collapsing rules when it substitutes the deduced template arguments for the parameter types. For example, passing the local variable s1 to the print_type_and_value function causes the compiler to produce the following function signature:
print_type_and_value<string&>(string& && t)
The compiler uses reference collapsing rules to reduce the signature to the following:
print_type_and_value<string&>(string& t)
This version of the print_type_and_value function then forwards its parameter to the correct specialized version of the S::print method.
The following table summarizes the reference collapsing rules for template argument type deduction:
Expanded type |
Collapsed type |
T& & |
T& |
T& && |
T& |
T&& & |
T& |
T&& && |
T&& |
Template argument deduction is an important element of implementing perfect forwarding. The section Perfect Forwarding, which is presented earlier in this topic, describes perfect forwarding in more detail.
Summary
Rvalue references distinguish lvalues from rvalues. They can help you improve the performance of your applications by eliminating the need for unnecessary memory allocations and copy operations. They also enable you to write one version of a function that accepts arbitrary arguments and forwards them to another function as if the other function had been called directly.
See Also
Tasks
How to: Write a Move Constructor
Reference
Expressions with Unary Operators
Lvalue Reference Declarator: &