How can I lock a mutex in an initializer list? - c++

I have a ConcurrentQueue class that is based around a user provided container with a constructor like this...
ConcurrentQueue(const ConcurrentQueue& other) : m_Queue(other.m_Queue) {}
But, I need to lock other's mutex while it's being copied.
Option 1:
So I could not use the copy constructor at all, and do...
ConcurrentQueue(const ConcurrentQueue& other) : m_Queue(other.m_Queue)
{
std::lock_guard<std::mutex> lock(other.m_Mutex);
m_Queue = other.m_Queue;
}
But I can't guarantee that copy assignment and copy construction are equivalent functionality.
Option 2:
I could have a private method...
std::queue<T, Container> GetQueue() const
{
std::lock_guard<std::mutex> lock(other.m_Mutex);
return m_Queue;
}
And then in the constructor do this...
ConcurrentQueue(const ConcurrentQueue& other) : m_Queue(other.GetQueue()) {}
But this potentially (depending on optimizations) uses m_Queue's copy constructor once and it's move constructor once. And I also can't guarantee that a copy and a move is equivalent to just a copy. Additionally the user provided container could be bizarre and be copyable but unmoveable, which would also cause this approach to have problems.
So, what do I do?

ConcurrrentQueue::ConcurrrentQueue(
ConcurrrentQueue const& other )
: m_Queue( (std::lock_guard<std::mutex>( other.m_Mutex ),
other.m_Queue ) )
{
}
should work.

Lock, create a copy of the content and then swap it with the member. At least that's the easiest and IMHO cleanest way. Another less clean way is to use the comma operator: (a, b) yields b, but if a is a scoped lock, the temporary will live until the next sequence point, i.e. until you have used b to initialize your local copy.
That said, there are two things to consider:
Maybe copying is not such a smart idea anyway and your design works as well if you just disable copying.
If you have access to the queue and you can read it for copying, doesn't that imply that the mutex must already be locked? If not, how are you sure that you really want to copy the queue? I don't question that there are answers that justify the design, but it's unusual.

Related

Why not to copy a class consisting of a single shared_ptr

Recently, I read an article about lazy data structures in C++ (this question is not about lazy, or the particular data structure, though - it is just the motivation).
A lazy stream (list) is implemented as follows:
template<class T>
class Stream
{
private:
std::shared_ptr <Susp<Cell<T>>> _lazyCell;
public:
Stream() {}
Stream(std::function<Cell<T>()> f)
: _lazyCell(std::make_shared<Susp<Cell<T>>>(f))
{}
Stream(Stream && stm)
: _lazyCell(std::move(stm._lazyCell))
{}
Stream & operator=(Stream && stm)
{
_lazyCell = std::move(stm._lazyCell);
return *this;
}
bool isEmpty() const
{
return !_lazyCell;
}
T get() const
{
return _lazyCell->get().val();
}
Stream<T> pop_front() const
{
return _lazyCell->get().pop_front();
}
};
The author mentions the move constructor:
I also added a move constructor and a move assignment operator for efficiency.
However, due to the explicit presence, one cannot simply assign a Stream. What is the motivation behind this?
As far as I can tell, the class consists solely of a shared_ptr, which can be trivially copied. Is there any benefit in forbidding copy-construction in such a class?
The shared_ptr is used internally to share lazy value cells as part of the private implementation.
However, from the user's point of view, it's an immutable object. Providing a copy constructor and assignment operator would undo this immutability.
He is modelling Haskell's immutable object's behaviour.
If it were thread-safe to do so, it would be reasonable to make this object copyable since in reality it's a handle to an (albeit more complex than usual) shared impl.
However, copyers would need to understand that they were copying a handle to shared state, and not state itself.
I think this is a type of premature optimization.
First of all, due to the rule of three/five (http://en.cppreference.com/w/cpp/language/rule_of_three), the move copy/assignment constructors would have been created in the same way without typing it out.
Therefore the only difference is, as you already pointed out, the missing copy/assigment constructor. It would have been better practice to mark them as deleted, e.g.:
Stream(Stream & stm) = deleted;
However the real problem here is using a shared pointer with only a single owner. It would have been much better to just use a std::unique_ptr. With its use copy and assignment are disabled automatically and the intent of the author is much clearer.

Why do I need the copy-and-swap idiom? [duplicate]

What is the copy-and-swap idiom and when should it be used? What problems does it solve? Does it change for C++11?
Related:
What are your favorite C++ Coding Style idioms: Copy-swap
Copy constructor and = operator overload in C++: is a common function possible?
What is copy elision and how it optimizes copy-and-swap idiom
C++: dynamically allocating an array of objects?
Overview
Why do we need the copy-and-swap idiom?
Any class that manages a resource (a wrapper, like a smart pointer) needs to implement The Big Three. While the goals and implementation of the copy-constructor and destructor are straightforward, the copy-assignment operator is arguably the most nuanced and difficult. How should it be done? What pitfalls need to be avoided?
The copy-and-swap idiom is the solution, and elegantly assists the assignment operator in achieving two things: avoiding code duplication, and providing a strong exception guarantee.
How does it work?
Conceptually, it works by using the copy-constructor's functionality to create a local copy of the data, then takes the copied data with a swap function, swapping the old data with the new data. The temporary copy then destructs, taking the old data with it. We are left with a copy of the new data.
In order to use the copy-and-swap idiom, we need three things: a working copy-constructor, a working destructor (both are the basis of any wrapper, so should be complete anyway), and a swap function.
A swap function is a non-throwing function that swaps two objects of a class, member for member. We might be tempted to use std::swap instead of providing our own, but this would be impossible; std::swap uses the copy-constructor and copy-assignment operator within its implementation, and we'd ultimately be trying to define the assignment operator in terms of itself!
(Not only that, but unqualified calls to swap will use our custom swap operator, skipping over the unnecessary construction and destruction of our class that std::swap would entail.)
An in-depth explanation
The goal
Let's consider a concrete case. We want to manage, in an otherwise useless class, a dynamic array. We start with a working constructor, copy-constructor, and destructor:
#include <algorithm> // std::copy
#include <cstddef> // std::size_t
class dumb_array
{
public:
// (default) constructor
dumb_array(std::size_t size = 0)
: mSize(size),
mArray(mSize ? new int[mSize]() : nullptr)
{
}
// copy-constructor
dumb_array(const dumb_array& other)
: mSize(other.mSize),
mArray(mSize ? new int[mSize] : nullptr)
{
// note that this is non-throwing, because of the data
// types being used; more attention to detail with regards
// to exceptions must be given in a more general case, however
std::copy(other.mArray, other.mArray + mSize, mArray);
}
// destructor
~dumb_array()
{
delete [] mArray;
}
private:
std::size_t mSize;
int* mArray;
};
This class almost manages the array successfully, but it needs operator= to work correctly.
A failed solution
Here's how a naive implementation might look:
// the hard part
dumb_array& operator=(const dumb_array& other)
{
if (this != &other) // (1)
{
// get rid of the old data...
delete [] mArray; // (2)
mArray = nullptr; // (2) *(see footnote for rationale)
// ...and put in the new
mSize = other.mSize; // (3)
mArray = mSize ? new int[mSize] : nullptr; // (3)
std::copy(other.mArray, other.mArray + mSize, mArray); // (3)
}
return *this;
}
And we say we're finished; this now manages an array, without leaks. However, it suffers from three problems, marked sequentially in the code as (n).
The first is the self-assignment test.
This check serves two purposes: it's an easy way to prevent us from running needless code on self-assignment, and it protects us from subtle bugs (such as deleting the array only to try and copy it). But in all other cases it merely serves to slow the program down, and act as noise in the code; self-assignment rarely occurs, so most of the time this check is a waste.
It would be better if the operator could work properly without it.
The second is that it only provides a basic exception guarantee. If new int[mSize] fails, *this will have been modified. (Namely, the size is wrong and the data is gone!)
For a strong exception guarantee, it would need to be something akin to:
dumb_array& operator=(const dumb_array& other)
{
if (this != &other) // (1)
{
// get the new data ready before we replace the old
std::size_t newSize = other.mSize;
int* newArray = newSize ? new int[newSize]() : nullptr; // (3)
std::copy(other.mArray, other.mArray + newSize, newArray); // (3)
// replace the old data (all are non-throwing)
delete [] mArray;
mSize = newSize;
mArray = newArray;
}
return *this;
}
The code has expanded! Which leads us to the third problem: code duplication.
Our assignment operator effectively duplicates all the code we've already written elsewhere, and that's a terrible thing.
In our case, the core of it is only two lines (the allocation and the copy), but with more complex resources this code bloat can be quite a hassle. We should strive to never repeat ourselves.
(One might wonder: if this much code is needed to manage one resource correctly, what if my class manages more than one?
While this may seem to be a valid concern, and indeed it requires non-trivial try/catch clauses, this is a non-issue.
That's because a class should manage one resource only!)
A successful solution
As mentioned, the copy-and-swap idiom will fix all these issues. But right now, we have all the requirements except one: a swap function. While The Rule of Three successfully entails the existence of our copy-constructor, assignment operator, and destructor, it should really be called "The Big Three and A Half": any time your class manages a resource it also makes sense to provide a swap function.
We need to add swap functionality to our class, and we do that as follows†:
class dumb_array
{
public:
// ...
friend void swap(dumb_array& first, dumb_array& second) // nothrow
{
// enable ADL (not necessary in our case, but good practice)
using std::swap;
// by swapping the members of two objects,
// the two objects are effectively swapped
swap(first.mSize, second.mSize);
swap(first.mArray, second.mArray);
}
// ...
};
(Here is the explanation why public friend swap.) Now not only can we swap our dumb_array's, but swaps in general can be more efficient; it merely swaps pointers and sizes, rather than allocating and copying entire arrays. Aside from this bonus in functionality and efficiency, we are now ready to implement the copy-and-swap idiom.
Without further ado, our assignment operator is:
dumb_array& operator=(dumb_array other) // (1)
{
swap(*this, other); // (2)
return *this;
}
And that's it! With one fell swoop, all three problems are elegantly tackled at once.
Why does it work?
We first notice an important choice: the parameter argument is taken by-value. While one could just as easily do the following (and indeed, many naive implementations of the idiom do):
dumb_array& operator=(const dumb_array& other)
{
dumb_array temp(other);
swap(*this, temp);
return *this;
}
We lose an important optimization opportunity. Not only that, but this choice is critical in C++11, which is discussed later. (On a general note, a remarkably useful guideline is as follows: if you're going to make a copy of something in a function, let the compiler do it in the parameter list.‡)
Either way, this method of obtaining our resource is the key to eliminating code duplication: we get to use the code from the copy-constructor to make the copy, and never need to repeat any bit of it. Now that the copy is made, we are ready to swap.
Observe that upon entering the function that all the new data is already allocated, copied, and ready to be used. This is what gives us a strong exception guarantee for free: we won't even enter the function if construction of the copy fails, and it's therefore not possible to alter the state of *this. (What we did manually before for a strong exception guarantee, the compiler is doing for us now; how kind.)
At this point we are home-free, because swap is non-throwing. We swap our current data with the copied data, safely altering our state, and the old data gets put into the temporary. The old data is then released when the function returns. (Where upon the parameter's scope ends and its destructor is called.)
Because the idiom repeats no code, we cannot introduce bugs within the operator. Note that this means we are rid of the need for a self-assignment check, allowing a single uniform implementation of operator=. (Additionally, we no longer have a performance penalty on non-self-assignments.)
And that is the copy-and-swap idiom.
What about C++11?
The next version of C++, C++11, makes one very important change to how we manage resources: the Rule of Three is now The Rule of Four (and a half). Why? Because not only do we need to be able to copy-construct our resource, we need to move-construct it as well.
Luckily for us, this is easy:
class dumb_array
{
public:
// ...
// move constructor
dumb_array(dumb_array&& other) noexcept ††
: dumb_array() // initialize via default constructor, C++11 only
{
swap(*this, other);
}
// ...
};
What's going on here? Recall the goal of move-construction: to take the resources from another instance of the class, leaving it in a state guaranteed to be assignable and destructible.
So what we've done is simple: initialize via the default constructor (a C++11 feature), then swap with other; we know a default constructed instance of our class can safely be assigned and destructed, so we know other will be able to do the same, after swapping.
(Note that some compilers do not support constructor delegation; in this case, we have to manually default construct the class. This is an unfortunate but luckily trivial task.)
Why does that work?
That is the only change we need to make to our class, so why does it work? Remember the ever-important decision we made to make the parameter a value and not a reference:
dumb_array& operator=(dumb_array other); // (1)
Now, if other is being initialized with an rvalue, it will be move-constructed. Perfect. In the same way C++03 let us re-use our copy-constructor functionality by taking the argument by-value, C++11 will automatically pick the move-constructor when appropriate as well. (And, of course, as mentioned in previously linked article, the copying/moving of the value may simply be elided altogether.)
And so concludes the copy-and-swap idiom.
Footnotes
*Why do we set mArray to null? Because if any further code in the operator throws, the destructor of dumb_array might be called; and if that happens without setting it to null, we attempt to delete memory that's already been deleted! We avoid this by setting it to null, as deleting null is a no-operation.
†There are other claims that we should specialize std::swap for our type, provide an in-class swap along-side a free-function swap, etc. But this is all unnecessary: any proper use of swap will be through an unqualified call, and our function will be found through ADL. One function will do.
‡The reason is simple: once you have the resource to yourself, you may swap and/or move it (C++11) anywhere it needs to be. And by making the copy in the parameter list, you maximize optimization.
††The move constructor should generally be noexcept, otherwise some code (e.g. std::vector resizing logic) will use the copy constructor even when a move would make sense. Of course, only mark it noexcept if the code inside doesn't throw exceptions.
Assignment, at its heart, is two steps: tearing down the object's old state and building its new state as a copy of some other object's state.
Basically, that's what the destructor and the copy constructor do, so the first idea would be to delegate the work to them. However, since destruction mustn't fail, while construction might, we actually want to do it the other way around: first perform the constructive part and, if that succeeded, then do the destructive part. The copy-and-swap idiom is a way to do just that: It first calls a class' copy constructor to create a temporary object, then swaps its data with the temporary's, and then lets the temporary's destructor destroy the old state.
Since swap() is supposed to never fail, the only part which might fail is the copy-construction. That is performed first, and if it fails, nothing will be changed in the targeted object.
In its refined form, copy-and-swap is implemented by having the copy performed by initializing the (non-reference) parameter of the assignment operator:
T& operator=(T tmp)
{
this->swap(tmp);
return *this;
}
There are some good answers already. I'll focus mainly on what I think they lack - an explanation of the "cons" with the copy-and-swap idiom....
What is the copy-and-swap idiom?
A way of implementing the assignment operator in terms of a swap function:
X& operator=(X rhs)
{
swap(rhs);
return *this;
}
The fundamental idea is that:
the most error-prone part of assigning to an object is ensuring any resources the new state needs are acquired (e.g. memory, descriptors)
that acquisition can be attempted before modifying the current state of the object (i.e. *this) if a copy of the new value is made, which is why rhs is accepted by value (i.e. copied) rather than by reference
swapping the state of the local copy rhs and *this is usually relatively easy to do without potential failure/exceptions, given the local copy doesn't need any particular state afterwards (just needs state fit for the destructor to run, much as for an object being moved from in >= C++11)
When should it be used? (Which problems does it solve [/create]?)
When you want the assigned-to objected unaffected by an assignment that throws an exception, assuming you have or can write a swap with strong exception guarantee, and ideally one that can't fail/throw..†
When you want a clean, easy to understand, robust way to define the assignment operator in terms of (simpler) copy constructor, swap and destructor functions.
Self-assignment done as a copy-and-swap avoids oft-overlooked edge cases.‡
When any performance penalty or momentarily higher resource usage created by having an extra temporary object during the assignment is not important to your application. ⁂
† swap throwing: it's generally possible to reliably swap data members that the objects track by pointer, but non-pointer data members that don't have a throw-free swap, or for which swapping has to be implemented as X tmp = lhs; lhs = rhs; rhs = tmp; and copy-construction or assignment may throw, still have the potential to fail leaving some data members swapped and others not. This potential applies even to C++03 std::string's as James comments on another answer:
#wilhelmtell: In C++03, there is no mention of exceptions potentially thrown by std::string::swap (which is called by std::swap). In C++0x, std::string::swap is noexcept and must not throw exceptions. – James McNellis Dec 22 '10 at 15:24
‡ assignment operator implementation that seems sane when assigning from a distinct object can easily fail for self-assignment. While it might seem unimaginable that client code would even attempt self-assignment, it can happen relatively easily during algo operations on containers, with x = f(x); code where f is (perhaps only for some #ifdef branches) a macro ala #define f(x) x or a function returning a reference to x, or even (likely inefficient but concise) code like x = c1 ? x * 2 : c2 ? x / 2 : x;). For example:
struct X
{
T* p_;
size_t size_;
X& operator=(const X& rhs)
{
delete[] p_; // OUCH!
p_ = new T[size_ = rhs.size_];
std::copy(p_, rhs.p_, rhs.p_ + rhs.size_);
}
...
};
On self-assignment, the above code delete's x.p_;, points p_ at a newly allocated heap region, then attempts to read the uninitialised data therein (Undefined Behaviour), if that doesn't do anything too weird, copy attempts a self-assignment to every just-destructed 'T'!
⁂ The copy-and-swap idiom can introduce inefficiencies or limitations due to the use of an extra temporary (when the operator's parameter is copy-constructed):
struct Client
{
IP_Address ip_address_;
int socket_;
X(const X& rhs)
: ip_address_(rhs.ip_address_), socket_(connect(rhs.ip_address_))
{ }
};
Here, a hand-written Client::operator= might check if *this is already connected to the same server as rhs (perhaps sending a "reset" code if useful), whereas the copy-and-swap approach would invoke the copy-constructor which would likely be written to open a distinct socket connection then close the original one. Not only could that mean a remote network interaction instead of a simple in-process variable copy, it could run afoul of client or server limits on socket resources or connections. (Of course this class has a pretty horrid interface, but that's another matter ;-P).
This answer is more like an addition and a slight modification to the answers above.
In some versions of Visual Studio (and possibly other compilers) there is a bug that is really annoying and doesn't make sense. So if you declare/define your swap function like this:
friend void swap(A& first, A& second) {
std::swap(first.size, second.size);
std::swap(first.arr, second.arr);
}
... the compiler will yell at you when you call the swap function:
This has something to do with a friend function being called and this object being passed as a parameter.
A way around this is to not use friend keyword and redefine the swap function:
void swap(A& other) {
std::swap(size, other.size);
std::swap(arr, other.arr);
}
This time, you can just call swap and pass in other, thus making the compiler happy:
After all, you don't need to use a friend function to swap 2 objects. It makes just as much sense to make swap a member function that has one other object as a parameter.
You already have access to this object, so passing it in as a parameter is technically redundant.
I would like to add a word of warning when you are dealing with C++11-style allocator-aware containers. Swapping and assignment have subtly different semantics.
For concreteness, let us consider a container std::vector<T, A>, where A is some stateful allocator type, and we'll compare the following functions:
void fs(std::vector<T, A> & a, std::vector<T, A> & b)
{
a.swap(b);
b.clear(); // not important what you do with b
}
void fm(std::vector<T, A> & a, std::vector<T, A> & b)
{
a = std::move(b);
}
The purpose of both functions fs and fm is to give a the state that b had initially. However, there is a hidden question: What happens if a.get_allocator() != b.get_allocator()? The answer is: It depends. Let's write AT = std::allocator_traits<A>.
If AT::propagate_on_container_move_assignment is std::true_type, then fm reassigns the allocator of a with the value of b.get_allocator(), otherwise it does not, and a continues to use its original allocator. In that case, the data elements need to be swapped individually, since the storage of a and b is not compatible.
If AT::propagate_on_container_swap is std::true_type, then fs swaps both data and allocators in the expected fashion.
If AT::propagate_on_container_swap is std::false_type, then we need a dynamic check.
If a.get_allocator() == b.get_allocator(), then the two containers use compatible storage, and swapping proceeds in the usual fashion.
However, if a.get_allocator() != b.get_allocator(), the program has undefined behaviour (cf. [container.requirements.general/8].
The upshot is that swapping has become a non-trivial operation in C++11 as soon as your container starts supporting stateful allocators. That's a somewhat "advanced use case", but it's not entirely unlikely, since move optimizations usually only become interesting once your class manages a resource, and memory is one of the most popular resources.

Adding a move only type to a class makes that class as a move only type?

In Scott Meyers' "Effective Modern C++", he writes:
It's worth nothing that because std::mutex is a move-only type (i.e., a type that can be
moved, but not copied), a side effect of adding m to Polynomial is that Polynomial loses the ability to be copied. It can still be moved, however.
Here, he adds a std::mutex m as a member variable to a class Polynomial in the context of explaining why one should protect access to const member variable of a class(while multi-threading). I understood the concept he explained. but one thing which I need to have some more explanation is "why Adding a move only type to a class makes that class as move only type not copy-able"?, why types like std::mutex and std::atomic is move only type by default? how can we get over with it if we want to perform copy operation even when adding move only type of variable to our class?
"why Adding a move only type to a class makes that class as move only type not copy-able"
This is not entierly true. adding non copyable member variable prevents you from writing something like Type (const Type& t) = default; , but you can still implement you own copy constructor which ignores the non copyable member
class Type{
int integer;
std::mutex m;
public:
Type() = default;
Type(const Type& rhs): integer (rhs.integer){}
}
Other than this, you can find numerous examples of objects that contain move-only members, but they are themselves non movable, simply because the developer deleted the move constructor.
why types like std::mutex and std::atomic is move only type by default?
Neither of them implement move constructor. this is a wrong statement.
std::mutex is not copyable only because the standard says so. They could probably have written a std::mutex that was copyable, but they decided not to.
As a guess, a non-copyable std::mutex might be more efficient, safer, or easier to understand.
As an example, here is why it might be safer/easier to understand: What does it mean to copy a std::mutex that is locked? I, personally, have no idea what the right answer is; I suspect that the right answer is "that is nonsense". There is no "answer of least surprise".
Instead of having a surprising answer in concurrency code, by blocking copy we avoid the question. People storing a mutex who want to copy have to decide themselves what they want it to do.
C++ will under some circumstances automatically generate a copy constructor. Under more circumstances it will let you do MyClass(MyClass const&)=default and write one when you ask for it.
If your class contains a non-copyable class, it won't generate one, and you cannot ask for it. This is because the generated copy constructor basically copies its members; if you cannot copy a member, it cannot generate a copy constructor.
So mutex cannot be copied because the standard says so.
If a struct or class contains a non-copyable member, the default copy constructor cannot be used. You have to write one yourself explicitly.
If your type needs to store a mutex, one approach to prevent the headache of having to maintain a copy ctor that manually copies all of its other members is to stick your non-mutex state into a sub-struct, and then use its default copy. Then you decide what to do with the mutex during the copy, and your copy ctor (and assignment copy operator) remain simple.
struct bob_with_mutex {
sturct bob_simple_data {
int x, y, z;
std::vector<char> buff;
};
bob_simple_data data;
std::mutex m;
bob_with_mutex( bob_with_mutex const& o ):
data(o.data)
{}
bob_with_mutex& operator=( bob_with_mutex const& o )
{
data = o.data;
return *this;
}
};
the type bob has a mixture of a mutex and some data. The data is stored in a sub-structure. The copy ctor of bob_with_mutex cannot be defaulted, but it simply says "copy the data, and nothing else".
The copy of the bob has its own mutex.
One possibility is to change the mutex type to a pointer to a mutex. You can copy the pointer, but there will still only be one copy of the mutex. You may want to use a smart pointer.

How to deal with noncopyable objects when inserting to containers in C++

I'm looking for the best-practice of dealing with non-copyable objects.
I have a mutex class, that obviously should not be copyable.
I added a private copy constructor to enforce that.
That broke the code - some places simply needed to be fixed, but I have a generic problem
where a class, using the mutex either as a data member, or by inheritance, is being inserted into a container.
This is usually happening during the container initialization, so the mutex is not initialized yet, and is therefore ok, but without a copy constructor it does not work. Changing the containers to contain pointers is not acceptable.
Any advise?
Three solutions here:
1. Use Pointers - The quick fix is to make it a container of pointers - e.g. a shared_ptr.
That would be the "good" solution if your objects are truly noncopyable, and using other containers is not possible.
2. Other containers - Alternatively, you could use non-copying containers (that use in-place-construction), however they aren't very common and largely incompatible with STL. (I've tried for a while, but it's simply no good)
That would be the "god" solution if your objects are truly noncopyable, and using pointers is not possible.
[edit] With C++13, std::vector allows inplace construction (emplace_back), and can be used for noncopyable objects that do implement move semantics.
[/edit]
3. Fix your copyability - If your class is copyable as such, and the mutex is not, you "simply" need to fix the copy constructor and assignment operator.
Writing them is a pain, since you usually have to copy & assign all members except the mutex, but that can often be simplified by:
template <typename TNonCopyable>
struct NeverCopy : public T
{
NeverCopy() {}
NeverCopy(T const & rhs) {}
NeverCopy<T> & operator=(T const & rhs) { return *this; }
}
And changing you mutex member to
NeverCopy<Mutex> m_mutex;
Unfortunately, using that template you lose special constructors of Mutex.
[edit] Warning: "Fixing" the Copy CTor/asignment often requires you to lock the right hand side on copy construct, and lock both sides on assignment. Unfortunately, there is no way to override the copy ctor/assignment and call the default implementation, so the NeverCopy trick might not work for you without external locking. (There are some other workarounds with their own limitations.)
If they are non-copyable, the container has to store (smart) pointers to those objects, or reference wrappers, etc, although with C++0x, noncopyable objects can still be movable (like boost threads), so that they can be stored in containers as-is.
to give examples: reference wrapper (aka boost::ref, a pointer under the hood)
#include <vector>
#include <tr1/functional>
struct Noncopy {
private:
Noncopy(const Noncopy&) {}
public:
Noncopy() {}
};
int main()
{
std::vector<std::tr1::reference_wrapper<Noncopy> > v;
Noncopy m;
v.push_back(std::tr1::reference_wrapper<Noncopy>(m));
}
C++0x, tested with gcc:
#include <vector>
struct Movable {
private:
Movable(const Movable&) = delete;
public:
Movable() {}
Movable(Movable&&) {}
};
int main()
{
std::vector<Movable> v;
Movable m;
v.emplace_back(std::move(m));
}
EDIT: Nevermind, C++0x FCD says, under 30.4.1/3,
A Mutex type shall not be copyable nor movable.
So you're better off with pointers to them. Smart or otherwise wrapped as necessary.
If an object is non-copyable then there is usually a good reason. And if there's a good reason then you shouldnt subvert that by putting it into a container that attempts to copy it.
There is no real answer to the question given how you've framed it. There is no way to do what you want. The actual answer is for the container to contain pointers, and you've said that isn't OK for some unspecified reason.
Some have talked about things being movable and using C++0x in which containers often require their elements to be movable, but do not require them to be copyable. I think this is a poor solution as well because I suspect that mutexes should not be moved while they are held, and this makes it effectively impossible to move them.
So, the only real remaining answer is to point at the mutexes. Use ::std::tr1::shared_ptr (in #include <tr1/memory>) or ::boost::shared_ptr to point at the mutexes. This requires changing the definitions of the classes that have the mutexes inside them, but it sounds like you're doing that anyway.
STL containers rely heavily on their contents being copyable, so either make them copyable or do not put them into container.
The best option is to use pointers or some type of wrapper class that uses pointers under the hood. That would allow these to be sanely copied, and actually do what a copy would be expected to do (share the mutex).
But, since you said no pointers, there is one other option. It sounds like Mutexes are "sometimes copyable" perhaps you should write a copy constructor and an assignment operator and have those throw an exception if a mutex is ever copied after it has been initialized. The down side is there's no way to know you're doing it wrong until runtime.
Use smart pointers like boost::shared_ptr or use another containers, like boost::intrusive. Both will require to modify your code, through.
Using a mutex in a class does not necessarily mean that the class has to be non-copyable. You can (almost) always implement it like this:
C::C (C const & c)
// No ctor-initializer here.
{
MutexLock guard (c.mutex);
// Do the copy-construction here.
x = c.x;
}
While this makes it somewhat possible to copy classes with mutexes, you probably should not do it. Chances are that your design will be better without per-instance mutex.
Using c++11 on Ubuntu 14.04 (which includes emplace_back), I've gotten this to work.
I found that emplace_back worked fine, but erase (and probably insert) didn't work because, when the vector was shuffling the elements along to fill in the gap, it mas using:
*previous = *current;
I found the trick was allowing move assignment in my resource class:
Watch& operator=(Watch &&other);
This is my inotify_watch class, which can live in a std::vector:
class Watch {
private:
int inotify_handle = 0;
int handle = -1;
// Erases all knowledge of our resources
void erase() {
inotify_handle = 0;
handle = -1;
}
public:
Watch(int inotify_handle, const char *path, uint32_t mask)
: inotify_handle(inotify_handle),
handle(inotify_add_watch(inotify_handle, path, mask)) {
if (handle == -1)
throw std::system_error(errno, std::system_category());
}
Watch(const Watch& other) = delete; // Can't copy it, it's a real resource
// Move is fine
Watch(Watch &&other)
: inotify_handle(other.inotify_handle), handle(other.handle) {
other.erase(); // Make the other one forget about our resources, so that
// when the destructor is called, it won't try to free them,
// as we own them now
}
// Move assignment is fine
Watch &operator=(Watch &&other) {
inotify_handle = other.inotify_handle;
handle = other.handle;
other.erase(); // Make the other one forget about our resources, so that
// when the destructor is called, it won't try to free them,
// as we own them now
return *this;
}
bool operator ==(const Watch& other) {
return (inotify_handle == other.inotify_handle) && (handle == other.handle);
}
~Watch() {
if (handle != -1) {
int result = inotify_rm_watch(inotify_handle, handle);
if (result == -1)
throw std::system_error(errno, std::system_category());
}
}
};
std::vector can not store non-copyable objects (due to resize) thus you can't store objects of type Foo:
struct Foo {
std::mutex mutex;
...
};
One way around this is to use std::unique_ptr:
struct Foo {
std::unique_ptr<std::mutex> pmutex;
Foo() : pmutex{std::make_unique<std::mutex>()} {}
...
};
Another option is to use a std::deque which can hold non-copyable objects (like instances of the first version of Foo above. Typically use emplace_back method to construct objects "in place" to add elements -- no copies happen. For example,
objects of type Foo here do not need to be copiable:
struct FooPool {
std::deque<Foo> objects;
ObjectPool(std::initializer_list<T> argList) {
for (auto&& arg : argList)
objects.emplace_back(arg);
...
};

What is the copy-and-swap idiom?

What is the copy-and-swap idiom and when should it be used? What problems does it solve? Does it change for C++11?
Related:
What are your favorite C++ Coding Style idioms: Copy-swap
Copy constructor and = operator overload in C++: is a common function possible?
What is copy elision and how it optimizes copy-and-swap idiom
C++: dynamically allocating an array of objects?
Overview
Why do we need the copy-and-swap idiom?
Any class that manages a resource (a wrapper, like a smart pointer) needs to implement The Big Three. While the goals and implementation of the copy-constructor and destructor are straightforward, the copy-assignment operator is arguably the most nuanced and difficult. How should it be done? What pitfalls need to be avoided?
The copy-and-swap idiom is the solution, and elegantly assists the assignment operator in achieving two things: avoiding code duplication, and providing a strong exception guarantee.
How does it work?
Conceptually, it works by using the copy-constructor's functionality to create a local copy of the data, then takes the copied data with a swap function, swapping the old data with the new data. The temporary copy then destructs, taking the old data with it. We are left with a copy of the new data.
In order to use the copy-and-swap idiom, we need three things: a working copy-constructor, a working destructor (both are the basis of any wrapper, so should be complete anyway), and a swap function.
A swap function is a non-throwing function that swaps two objects of a class, member for member. We might be tempted to use std::swap instead of providing our own, but this would be impossible; std::swap uses the copy-constructor and copy-assignment operator within its implementation, and we'd ultimately be trying to define the assignment operator in terms of itself!
(Not only that, but unqualified calls to swap will use our custom swap operator, skipping over the unnecessary construction and destruction of our class that std::swap would entail.)
An in-depth explanation
The goal
Let's consider a concrete case. We want to manage, in an otherwise useless class, a dynamic array. We start with a working constructor, copy-constructor, and destructor:
#include <algorithm> // std::copy
#include <cstddef> // std::size_t
class dumb_array
{
public:
// (default) constructor
dumb_array(std::size_t size = 0)
: mSize(size),
mArray(mSize ? new int[mSize]() : nullptr)
{
}
// copy-constructor
dumb_array(const dumb_array& other)
: mSize(other.mSize),
mArray(mSize ? new int[mSize] : nullptr)
{
// note that this is non-throwing, because of the data
// types being used; more attention to detail with regards
// to exceptions must be given in a more general case, however
std::copy(other.mArray, other.mArray + mSize, mArray);
}
// destructor
~dumb_array()
{
delete [] mArray;
}
private:
std::size_t mSize;
int* mArray;
};
This class almost manages the array successfully, but it needs operator= to work correctly.
A failed solution
Here's how a naive implementation might look:
// the hard part
dumb_array& operator=(const dumb_array& other)
{
if (this != &other) // (1)
{
// get rid of the old data...
delete [] mArray; // (2)
mArray = nullptr; // (2) *(see footnote for rationale)
// ...and put in the new
mSize = other.mSize; // (3)
mArray = mSize ? new int[mSize] : nullptr; // (3)
std::copy(other.mArray, other.mArray + mSize, mArray); // (3)
}
return *this;
}
And we say we're finished; this now manages an array, without leaks. However, it suffers from three problems, marked sequentially in the code as (n).
The first is the self-assignment test.
This check serves two purposes: it's an easy way to prevent us from running needless code on self-assignment, and it protects us from subtle bugs (such as deleting the array only to try and copy it). But in all other cases it merely serves to slow the program down, and act as noise in the code; self-assignment rarely occurs, so most of the time this check is a waste.
It would be better if the operator could work properly without it.
The second is that it only provides a basic exception guarantee. If new int[mSize] fails, *this will have been modified. (Namely, the size is wrong and the data is gone!)
For a strong exception guarantee, it would need to be something akin to:
dumb_array& operator=(const dumb_array& other)
{
if (this != &other) // (1)
{
// get the new data ready before we replace the old
std::size_t newSize = other.mSize;
int* newArray = newSize ? new int[newSize]() : nullptr; // (3)
std::copy(other.mArray, other.mArray + newSize, newArray); // (3)
// replace the old data (all are non-throwing)
delete [] mArray;
mSize = newSize;
mArray = newArray;
}
return *this;
}
The code has expanded! Which leads us to the third problem: code duplication.
Our assignment operator effectively duplicates all the code we've already written elsewhere, and that's a terrible thing.
In our case, the core of it is only two lines (the allocation and the copy), but with more complex resources this code bloat can be quite a hassle. We should strive to never repeat ourselves.
(One might wonder: if this much code is needed to manage one resource correctly, what if my class manages more than one?
While this may seem to be a valid concern, and indeed it requires non-trivial try/catch clauses, this is a non-issue.
That's because a class should manage one resource only!)
A successful solution
As mentioned, the copy-and-swap idiom will fix all these issues. But right now, we have all the requirements except one: a swap function. While The Rule of Three successfully entails the existence of our copy-constructor, assignment operator, and destructor, it should really be called "The Big Three and A Half": any time your class manages a resource it also makes sense to provide a swap function.
We need to add swap functionality to our class, and we do that as follows†:
class dumb_array
{
public:
// ...
friend void swap(dumb_array& first, dumb_array& second) // nothrow
{
// enable ADL (not necessary in our case, but good practice)
using std::swap;
// by swapping the members of two objects,
// the two objects are effectively swapped
swap(first.mSize, second.mSize);
swap(first.mArray, second.mArray);
}
// ...
};
(Here is the explanation why public friend swap.) Now not only can we swap our dumb_array's, but swaps in general can be more efficient; it merely swaps pointers and sizes, rather than allocating and copying entire arrays. Aside from this bonus in functionality and efficiency, we are now ready to implement the copy-and-swap idiom.
Without further ado, our assignment operator is:
dumb_array& operator=(dumb_array other) // (1)
{
swap(*this, other); // (2)
return *this;
}
And that's it! With one fell swoop, all three problems are elegantly tackled at once.
Why does it work?
We first notice an important choice: the parameter argument is taken by-value. While one could just as easily do the following (and indeed, many naive implementations of the idiom do):
dumb_array& operator=(const dumb_array& other)
{
dumb_array temp(other);
swap(*this, temp);
return *this;
}
We lose an important optimization opportunity. Not only that, but this choice is critical in C++11, which is discussed later. (On a general note, a remarkably useful guideline is as follows: if you're going to make a copy of something in a function, let the compiler do it in the parameter list.‡)
Either way, this method of obtaining our resource is the key to eliminating code duplication: we get to use the code from the copy-constructor to make the copy, and never need to repeat any bit of it. Now that the copy is made, we are ready to swap.
Observe that upon entering the function that all the new data is already allocated, copied, and ready to be used. This is what gives us a strong exception guarantee for free: we won't even enter the function if construction of the copy fails, and it's therefore not possible to alter the state of *this. (What we did manually before for a strong exception guarantee, the compiler is doing for us now; how kind.)
At this point we are home-free, because swap is non-throwing. We swap our current data with the copied data, safely altering our state, and the old data gets put into the temporary. The old data is then released when the function returns. (Where upon the parameter's scope ends and its destructor is called.)
Because the idiom repeats no code, we cannot introduce bugs within the operator. Note that this means we are rid of the need for a self-assignment check, allowing a single uniform implementation of operator=. (Additionally, we no longer have a performance penalty on non-self-assignments.)
And that is the copy-and-swap idiom.
What about C++11?
The next version of C++, C++11, makes one very important change to how we manage resources: the Rule of Three is now The Rule of Four (and a half). Why? Because not only do we need to be able to copy-construct our resource, we need to move-construct it as well.
Luckily for us, this is easy:
class dumb_array
{
public:
// ...
// move constructor
dumb_array(dumb_array&& other) noexcept ††
: dumb_array() // initialize via default constructor, C++11 only
{
swap(*this, other);
}
// ...
};
What's going on here? Recall the goal of move-construction: to take the resources from another instance of the class, leaving it in a state guaranteed to be assignable and destructible.
So what we've done is simple: initialize via the default constructor (a C++11 feature), then swap with other; we know a default constructed instance of our class can safely be assigned and destructed, so we know other will be able to do the same, after swapping.
(Note that some compilers do not support constructor delegation; in this case, we have to manually default construct the class. This is an unfortunate but luckily trivial task.)
Why does that work?
That is the only change we need to make to our class, so why does it work? Remember the ever-important decision we made to make the parameter a value and not a reference:
dumb_array& operator=(dumb_array other); // (1)
Now, if other is being initialized with an rvalue, it will be move-constructed. Perfect. In the same way C++03 let us re-use our copy-constructor functionality by taking the argument by-value, C++11 will automatically pick the move-constructor when appropriate as well. (And, of course, as mentioned in previously linked article, the copying/moving of the value may simply be elided altogether.)
And so concludes the copy-and-swap idiom.
Footnotes
*Why do we set mArray to null? Because if any further code in the operator throws, the destructor of dumb_array might be called; and if that happens without setting it to null, we attempt to delete memory that's already been deleted! We avoid this by setting it to null, as deleting null is a no-operation.
†There are other claims that we should specialize std::swap for our type, provide an in-class swap along-side a free-function swap, etc. But this is all unnecessary: any proper use of swap will be through an unqualified call, and our function will be found through ADL. One function will do.
‡The reason is simple: once you have the resource to yourself, you may swap and/or move it (C++11) anywhere it needs to be. And by making the copy in the parameter list, you maximize optimization.
††The move constructor should generally be noexcept, otherwise some code (e.g. std::vector resizing logic) will use the copy constructor even when a move would make sense. Of course, only mark it noexcept if the code inside doesn't throw exceptions.
Assignment, at its heart, is two steps: tearing down the object's old state and building its new state as a copy of some other object's state.
Basically, that's what the destructor and the copy constructor do, so the first idea would be to delegate the work to them. However, since destruction mustn't fail, while construction might, we actually want to do it the other way around: first perform the constructive part and, if that succeeded, then do the destructive part. The copy-and-swap idiom is a way to do just that: It first calls a class' copy constructor to create a temporary object, then swaps its data with the temporary's, and then lets the temporary's destructor destroy the old state.
Since swap() is supposed to never fail, the only part which might fail is the copy-construction. That is performed first, and if it fails, nothing will be changed in the targeted object.
In its refined form, copy-and-swap is implemented by having the copy performed by initializing the (non-reference) parameter of the assignment operator:
T& operator=(T tmp)
{
this->swap(tmp);
return *this;
}
There are some good answers already. I'll focus mainly on what I think they lack - an explanation of the "cons" with the copy-and-swap idiom....
What is the copy-and-swap idiom?
A way of implementing the assignment operator in terms of a swap function:
X& operator=(X rhs)
{
swap(rhs);
return *this;
}
The fundamental idea is that:
the most error-prone part of assigning to an object is ensuring any resources the new state needs are acquired (e.g. memory, descriptors)
that acquisition can be attempted before modifying the current state of the object (i.e. *this) if a copy of the new value is made, which is why rhs is accepted by value (i.e. copied) rather than by reference
swapping the state of the local copy rhs and *this is usually relatively easy to do without potential failure/exceptions, given the local copy doesn't need any particular state afterwards (just needs state fit for the destructor to run, much as for an object being moved from in >= C++11)
When should it be used? (Which problems does it solve [/create]?)
When you want the assigned-to objected unaffected by an assignment that throws an exception, assuming you have or can write a swap with strong exception guarantee, and ideally one that can't fail/throw..†
When you want a clean, easy to understand, robust way to define the assignment operator in terms of (simpler) copy constructor, swap and destructor functions.
Self-assignment done as a copy-and-swap avoids oft-overlooked edge cases.‡
When any performance penalty or momentarily higher resource usage created by having an extra temporary object during the assignment is not important to your application. ⁂
† swap throwing: it's generally possible to reliably swap data members that the objects track by pointer, but non-pointer data members that don't have a throw-free swap, or for which swapping has to be implemented as X tmp = lhs; lhs = rhs; rhs = tmp; and copy-construction or assignment may throw, still have the potential to fail leaving some data members swapped and others not. This potential applies even to C++03 std::string's as James comments on another answer:
#wilhelmtell: In C++03, there is no mention of exceptions potentially thrown by std::string::swap (which is called by std::swap). In C++0x, std::string::swap is noexcept and must not throw exceptions. – James McNellis Dec 22 '10 at 15:24
‡ assignment operator implementation that seems sane when assigning from a distinct object can easily fail for self-assignment. While it might seem unimaginable that client code would even attempt self-assignment, it can happen relatively easily during algo operations on containers, with x = f(x); code where f is (perhaps only for some #ifdef branches) a macro ala #define f(x) x or a function returning a reference to x, or even (likely inefficient but concise) code like x = c1 ? x * 2 : c2 ? x / 2 : x;). For example:
struct X
{
T* p_;
size_t size_;
X& operator=(const X& rhs)
{
delete[] p_; // OUCH!
p_ = new T[size_ = rhs.size_];
std::copy(p_, rhs.p_, rhs.p_ + rhs.size_);
}
...
};
On self-assignment, the above code delete's x.p_;, points p_ at a newly allocated heap region, then attempts to read the uninitialised data therein (Undefined Behaviour), if that doesn't do anything too weird, copy attempts a self-assignment to every just-destructed 'T'!
⁂ The copy-and-swap idiom can introduce inefficiencies or limitations due to the use of an extra temporary (when the operator's parameter is copy-constructed):
struct Client
{
IP_Address ip_address_;
int socket_;
X(const X& rhs)
: ip_address_(rhs.ip_address_), socket_(connect(rhs.ip_address_))
{ }
};
Here, a hand-written Client::operator= might check if *this is already connected to the same server as rhs (perhaps sending a "reset" code if useful), whereas the copy-and-swap approach would invoke the copy-constructor which would likely be written to open a distinct socket connection then close the original one. Not only could that mean a remote network interaction instead of a simple in-process variable copy, it could run afoul of client or server limits on socket resources or connections. (Of course this class has a pretty horrid interface, but that's another matter ;-P).
This answer is more like an addition and a slight modification to the answers above.
In some versions of Visual Studio (and possibly other compilers) there is a bug that is really annoying and doesn't make sense. So if you declare/define your swap function like this:
friend void swap(A& first, A& second) {
std::swap(first.size, second.size);
std::swap(first.arr, second.arr);
}
... the compiler will yell at you when you call the swap function:
This has something to do with a friend function being called and this object being passed as a parameter.
A way around this is to not use friend keyword and redefine the swap function:
void swap(A& other) {
std::swap(size, other.size);
std::swap(arr, other.arr);
}
This time, you can just call swap and pass in other, thus making the compiler happy:
After all, you don't need to use a friend function to swap 2 objects. It makes just as much sense to make swap a member function that has one other object as a parameter.
You already have access to this object, so passing it in as a parameter is technically redundant.
I would like to add a word of warning when you are dealing with C++11-style allocator-aware containers. Swapping and assignment have subtly different semantics.
For concreteness, let us consider a container std::vector<T, A>, where A is some stateful allocator type, and we'll compare the following functions:
void fs(std::vector<T, A> & a, std::vector<T, A> & b)
{
a.swap(b);
b.clear(); // not important what you do with b
}
void fm(std::vector<T, A> & a, std::vector<T, A> & b)
{
a = std::move(b);
}
The purpose of both functions fs and fm is to give a the state that b had initially. However, there is a hidden question: What happens if a.get_allocator() != b.get_allocator()? The answer is: It depends. Let's write AT = std::allocator_traits<A>.
If AT::propagate_on_container_move_assignment is std::true_type, then fm reassigns the allocator of a with the value of b.get_allocator(), otherwise it does not, and a continues to use its original allocator. In that case, the data elements need to be swapped individually, since the storage of a and b is not compatible.
If AT::propagate_on_container_swap is std::true_type, then fs swaps both data and allocators in the expected fashion.
If AT::propagate_on_container_swap is std::false_type, then we need a dynamic check.
If a.get_allocator() == b.get_allocator(), then the two containers use compatible storage, and swapping proceeds in the usual fashion.
However, if a.get_allocator() != b.get_allocator(), the program has undefined behaviour (cf. [container.requirements.general/8].
The upshot is that swapping has become a non-trivial operation in C++11 as soon as your container starts supporting stateful allocators. That's a somewhat "advanced use case", but it's not entirely unlikely, since move optimizations usually only become interesting once your class manages a resource, and memory is one of the most popular resources.