c++ - What is the copy-and-swap idiom?

ID : 375

viewed : 235

Tags : c++copy-constructorassignment-operatorc++-faqcopy-and-swapc++

Top 5 Answer for c++ - What is the copy-and-swap idiom?

vote vote



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).

  1. 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.

  2. 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;  } 
  3. 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.


*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.

vote vote


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; } 
vote vote


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).

vote vote


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:

enter image description here

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:

enter image description here

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.

vote vote


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.

Top 3 video Explaining c++ - What is the copy-and-swap idiom?