gecko/mfbt/Move.h
Justin Lebar 2e2403726d Bug 909977 - Rename mozilla::Move to mozilla::OldMove, and make mozilla::Move a synonym for std::move(). r=waldo
--HG--
extra : rebase_source : 7b3bb02cc8cbc0ad6721c6c3895564d9567b8ddb
2013-08-29 11:54:14 -07:00

222 lines
7.7 KiB
C++

/* -*- Mode: C++; tab-width: 8; indent-tabs-mode: nil; c-basic-offset: 2 -*- */
/* vim: set ts=8 sts=2 et sw=2 tw=80: */
/* This Source Code Form is subject to the terms of the Mozilla Public
* License, v. 2.0. If a copy of the MPL was not distributed with this
* file, You can obtain one at http://mozilla.org/MPL/2.0/. */
/* C++11-style, but C++98-usable, "move references" implementation. */
#ifndef mozilla_Move_h
#define mozilla_Move_h
#include "mozilla/TypeTraits.h"
namespace mozilla {
/*
* "Move" References
*
* [Once upon a time, C++11 rvalue references were not implemented by all the
* compilers we cared about, so we invented mozilla::Move() (now called
* OldMove()), which does something similar. We're in the process of
* transitioning away from this to pure stl (bug 896100). Until that bug is
* completed, this header will provide both mozilla::OldMove() and
* mozilla::Move().]
*
*
* Some types can be copied much more efficiently if we know the original's
* value need not be preserved --- that is, if we are doing a "move", not a
* "copy". For example, if we have:
*
* Vector<T> u;
* Vector<T> v(u);
*
* the constructor for v must apply a copy constructor to each element of u ---
* taking time linear in the length of u. However, if we know we will not need u
* any more once v has been initialized, then we could initialize v very
* efficiently simply by stealing u's dynamically allocated buffer and giving it
* to v --- a constant-time operation, regardless of the size of u.
*
* Moves often appear in container implementations. For example, when we append
* to a vector, we may need to resize its buffer. This entails moving each of
* its extant elements from the old, smaller buffer to the new, larger buffer.
* But once the elements have been migrated, we're just going to throw away the
* old buffer; we don't care if they still have their values. So if the vector's
* element type can implement "move" more efficiently than "copy", the vector
* resizing should by all means use a "move" operation. Hash tables also need to
* be resized.
*
* The details of the optimization, and whether it's worth applying, vary from
* one type to the next. And while some constructor calls are moves, many really
* are copies, and can't be optimized this way. So we need:
*
* 1) a way for a particular invocation of a copy constructor to say that it's
* really a move, and that the value of the original isn't important
* afterwards (although it must still be safe to destroy); and
*
* 2) a way for a type (like Vector) to announce that it can be moved more
* efficiently than it can be copied, and provide an implementation of that
* move operation.
*
* The OldMove(T&) function takes a reference to a T, and returns a MoveRef<T>
* referring to the same value; that's (1). A MoveRef<T> is simply a reference
* to a T, annotated to say that a copy constructor applied to it may move that
* T, instead of copying it. Finally, a constructor that accepts an MoveRef<T>
* should perform a more efficient move, instead of a copy, providing (2).
*
* The Move(T&) function takes a reference to a T and returns a T&&. It acts
* just like std::move(), which is not available on all our platforms.
*
* In new code, you should use Move(T&) and T&& instead of OldMove(T&) and
* MoveRef<T>, where possible.
*
* Where we might define a copy constructor for a class C like this:
*
* C(const C& rhs) { ... copy rhs to this ... }
*
* we would declare a move constructor like this:
*
* C(C&& rhs) { .. move rhs to this ... }
*
* or, in the deprecated OldMove style:
*
* C(MoveRef<C> rhs) { ... move rhs to this ... }
*
* And where we might perform a copy like this:
*
* C c2(c1);
*
* we would perform a move like this:
*
* C c2(Move(c1));
*
* or, in the deprecated OldMove style:
*
* C c2(OldMove(c1));
*
* Note that MoveRef<T> implicitly converts to T&, so you can pass a MoveRef<T>
* to an ordinary copy constructor for a type that doesn't support a special
* move constructor, and you'll just get a copy. This means that templates can
* use Move whenever they know they won't use the original value any more, even
* if they're not sure whether the type at hand has a specialized move
* constructor. If it doesn't, the MoveRef<T> will just convert to a T&, and
* the ordinary copy constructor will apply.
*
* A class with a move constructor can also provide a move assignment operator,
* which runs this's destructor, and then applies the move constructor to
* *this's memory. A typical definition:
*
* C& operator=(C&& rhs) { // or |MoveRef<C> rhs|
* this->~C();
* new(this) C(rhs);
* return *this;
* }
*
* With that in place, one can write move assignments like this:
*
* c2 = Move(c1); // or OldMove()
*
* This destroys c1, moves c1's value to c2, and leaves c1 in an undefined but
* destructible state.
*
* This header file defines MoveRef, Move, and OldMove in the mozilla namespace.
* It's up to individual containers to annotate moves as such, by calling Move
* or OldMove; and it's up to individual types to define move constructors.
*
* One hint: if you're writing a move constructor where the type has members
* that should be moved themselves, it's much nicer to write this:
*
* C(MoveRef<C> c) : x(Move(c->x)), y(Move(c->y)) { }
*
* than the equivalent:
*
* C(MoveRef<C> c) { new(&x) X(Move(c->x)); new(&y) Y(Move(c->y)); }
*
* especially since GNU C++ fails to notice that this does indeed initialize x
* and y, which may matter if they're const.
*/
template<typename T>
class MoveRef
{
T* pointer;
public:
explicit MoveRef(T& t) : pointer(&t) { }
T& operator*() const { return *pointer; }
T* operator->() const { return pointer; }
operator T& () const { return *pointer; }
};
template<typename T>
inline MoveRef<T>
OldMove(T& t)
{
return MoveRef<T>(t);
}
template<typename T>
inline MoveRef<T>
OldMove(const T& t)
{
// With some versions of gcc, for a class C, there's an (incorrect) ambiguity
// between the C(const C&) constructor and the default C(C&&) C++11 move
// constructor, when the constructor is called with a const C& argument.
//
// This ambiguity manifests with the Move implementation above when Move is
// passed const U& for some class U. Calling Move(const U&) returns a
// MoveRef<const U&>, which is then commonly passed to the U constructor,
// triggering an implicit conversion to const U&. gcc doesn't know whether to
// call U(const U&) or U(U&&), so it wrongly reports a compile error.
//
// http://gcc.gnu.org/bugzilla/show_bug.cgi?id=50442 has since been fixed, so
// this is no longer an issue for up-to-date compilers. But there's no harm
// in keeping it around for older compilers, so we might as well. See also
// bug 686280.
return MoveRef<T>(const_cast<T&>(t));
}
/**
* Identical to std::Move(); this is necessary until our stlport supports
* std::move().
*/
template<typename T>
inline typename RemoveReference<T>::Type&&
Move(T&& a)
{
return static_cast<typename RemoveReference<T>::Type&&>(a);
}
/**
* These two overloads are identidal to std::Forward(); they are necessary until
* our stlport supports std::forward().
*/
template<typename T>
inline T&&
Forward(typename RemoveReference<T>::Type& a)
{
return static_cast<T&&>(a);
}
template<typename T>
inline T&&
Forward(typename RemoveReference<T>::Type&& t)
{
static_assert(!IsLvalueReference<T>::value,
"misuse of Forward detected! try the other overload");
return static_cast<T&&>(t);
}
/** Swap |t| and |u| using move-construction if possible. */
template<typename T>
inline void
Swap(T& t, T& u)
{
T tmp(OldMove(t));
t = OldMove(u);
u = OldMove(tmp);
}
} // namespace mozilla
#endif /* mozilla_Move_h */