[c++] What does T&& (double ampersand) mean in C++11?

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.

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What does T&& (double ampersand) mean in C++11?