C++ Double Address Operator? (&&)

This is C++11 code. In C++11, the && token can be used to mean an "rvalue reference".

What is && in c++

It is an rvalue reference. I could tell you that I am an expert on the matter, but this is probably the most confusing aspect of C++ to me. Anyhow, here is a great explanation.

http://thbecker.net/articles/rvalue_references/section_03.html

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

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);

c++ double & usage

This is a new thing in C++11 called rvalue references.

You can read a great introduction to them here: http://thbecker.net/articles/rvalue_references/section_01.html

Essentially, it says it will be a reference to an object which can be destroyed without causing problems, and is used to optimise certain operations such as copy constructors (which can be changed for a swap if the other object can be destroyed).

What does a && operator do when there is no left side in C?

It's a gcc-specific extension, a unary && operator that can be applied to a label name, yielding its address as a void* value.

As part of the extension, goto *ptr; is allowed where ptr is an expression of type void*.

It's documented here in the gcc manual.

You can get the address of a label defined in the current function (or
a containing function) with the unary operator &&. The value has
type void *. This value is a constant and can be used wherever a
constant of that type is valid. For example:
void *ptr;
/* ... */
ptr = &&foo;
To use these values, you need to be able to jump to one. This is done
with the computed goto statement, goto *exp;. For example,
goto *ptr;
Any expression of type void * is allowed.

As zwol points out in a comment, gcc uses && rather than the more obvious & because a label and an object with the same name can be visible simultaneously, making &foo potentially ambiguous if & means "address of label". Label names occupy their own namespace (not in the C++ sense), and can appear only in specific contexts: defined by a labeled-statement, as the target of a goto statement, or, for gcc, as the operand of unary &&.

What does the double ampersand return type mean?

From: http://www.stroustrup.com/C++11FAQ.html#rval

The && indicates an "rvalue reference". An rvalue reference can bind to an rvalue (but not to an lvalue):

X a;
X f();
X& r1 = a;      // bind r1 to a (an lvalue)
X& r2 = f();    // error: f() is an rvalue; can't bind

X&& rr1 = f();  // fine: bind rr1 to temporary
X&& rr2 = a;    // error: bind a is an lvalue

Assign a value to an rvalue reference returned from function

You said "When function calling finished, the object vector {1, 2, 3, 4, 5} will be destroyed" but that is untrue. The temporary created for the function call is not deleted until the statement ends, i.e. the next line of code. Otherwise imagine how much code would break that passes c_str() of a temporary string.

C++ override address operator

You can do it like this in C++11:

class Object
{
    public:
};

Object* operator&(Object&& object) { return std::addressof(object); }
Object const* operator&(Object const&& object) { return std::addressof(object); }

All objects:

template<typename T>
T* operator&(T&& x) {
    return std::addressof(x);
}

Please don't do this though. People will hate you for it.

C++ Double Address Operator? (&&)