In some cases, an instance of a C++ class should not be copied at all. There are three ways to prevent such an object copy: keeping the copy constructor and assignment operator private, using a special non-copyable mixin, or deleting those special member functions.
A class that represents a wrapper stream of a file should not have its instance copied around. It will cause a confusion in the handling of the actual I/O system. In a similar spirit, if an instance holds a unique private object, copying the pointer does not make sense. A somehow related problem but not necessarily similar is the issue of object slicing.
The following illustration demonstrates a simple class that is supposed to have a unique owner, an instance of .
For this purpose, the implementation of is as simple as:
To show the issue, a helper function is implement as follows:
From this example, it is obvious that an instance of must not be copied. In particular, another clone of a similar car should not automatically belong to the same owner. In fact, running the subsequent code:
will give the output:Owner is Joe Sixpack Owner is Joe Sixpack
How can we prevent this accidental object copy?
Method 1: Private copy constructor and copy assignment operator
A very common technique is to declare both the copy constructor and copy assignment operator to be private. We do not even need to implement them. The idea is so that any attempt to perform a copy or an assignment will provoke a compile error.
In the above example, will be modified to look like the following. Take a look closely at two additional private members of the class.
Now if we try again to assign an instance of to a new one, the compiler will complain loudly:example.cpp:35:22: error: calling a private constructor of class 'Car' Car anotherSedan = sedan; ^ example.cpp:22:3: note: declared private here Car(const Car&); ^ 1 error generated.
If writing two additional lines containing repetitive names is too cumbersome, a macro could be utilized instead. This is the approach used by WebKit, see its macro from wtf/Noncopyable.h (do not be alarmed, in the context of WebKit source code, WTF here stands for Web Template Framework). Chromium code, as shown in the file base/macros.h, distinguishes between copy constructor and assignment, denoted as and macros, respectively.
Method 2: Non-copyable mixin
The idea above can be extended to create a dedicated class which has the sole purpose to prevent object copying. It is often called as Noncopyable and typically used as a mixin. In our example, the class can then be derived from this .
Boost users may be already familiar with boost::noncopyable, the Boost flavor of the said mixin. A conceptual, self-contained implementation of that mixin will resemble something like the following:
Our lovely Car class can be written as:
Compared to the first method, using has the benefit of making the intention very clear. A quick glance at the class, right on its first line, and you know right away that its instance is not supposed to be copied.
Method 3: Deleted copy constructor and copy assignment operator
For modern applications, there is less and less reason to get stuck with the above workaround. Thanks to C++11, the solution becomes magically simple: just delete the copy constructor and assignment operator. Our class will look like this instead:
Note that if you use mixin with a compiler supporting C++11, the implementation of also automatically deletes the said member functions.
With this approach, any accidental copy will result in a quite friendlier error message:example.cpp:34:7: error: call to deleted constructor of 'Car' Car anotherSedan = sedan; ^ ~~~~~ example.cpp:10:3: note: 'Car' has been explicitly marked deleted here Car(const Car&) = delete; ^
So, which of the above three methods is your favorite?
A copy assignment operator of class is a non-template non-static member function with the name operator= that takes exactly one parameter of type T, T&, const T&, volatile T&, or constvolatile T&. For a type to be , it must have a public copy assignment operator.
|class_nameclass_name ( class_name )||(1)|
|class_nameclass_name ( const class_name )||(2)|
|class_nameclass_name ( const class_name ) = default;||(3)||(since C++11)|
|class_nameclass_name ( const class_name ) = delete;||(4)||(since C++11)|
- Typical declaration of a copy assignment operator when copy-and-swap idiom can be used.
- Typical declaration of a copy assignment operator when copy-and-swap idiom cannot be used (non-swappable type or degraded performance).
- Forcing a copy assignment operator to be generated by the compiler.
- Avoiding implicit copy assignment.
The copy assignment operator is called whenever selected by overload resolution, e.g. when an object appears on the left side of an assignment expression.
Implicitly-declared copy assignment operator
If no user-defined copy assignment operators are provided for a class type (struct, class, or union), the compiler will always declare one as an inline public member of the class. This implicitly-declared copy assignment operator has the form T& T::operator=(const T&) if all of the following is true:
- each direct base of has a copy assignment operator whose parameters are B or const B& or constvolatile B&;
- each non-static data member of of class type or array of class type has a copy assignment operator whose parameters are M or const M& or constvolatile M&.
Otherwise the implicitly-declared copy assignment operator is declared as T& T::operator=(T&). (Note that due to these rules, the implicitly-declared copy assignment operator cannot bind to a volatile lvalue argument.)
A class can have multiple copy assignment operators, e.g. both T& T::operator=(const T&) and T& T::operator=(T). If some user-defined copy assignment operators are present, the user may still force the generation of the implicitly declared copy assignment operator with the keyword .(since C++11)
The implicitly-declared (or defaulted on its first declaration) copy assignment operator has an exception specification as described in dynamic exception specification(until C++17)exception specification(since C++17)
Because the copy assignment operator is always declared for any class, the base class assignment operator is always hidden. If a using-declaration is used to bring in the assignment operator from the base class, and its argument type could be the same as the argument type of the implicit assignment operator of the derived class, the using-declaration is also hidden by the implicit declaration.
Deleted implicitly-declared copy assignment operator
A implicitly-declared copy assignment operator for class is defined as deleted if any of the following is true:
- has a user-declared move constructor;
- has a user-declared move assignment operator.
Otherwise, it is defined as defaulted.
A defaulted copy assignment operator for class is defined as deleted if any of the following is true:
- has a non-static data member of non-class type (or array thereof) that is const;
- has a non-static data member of a reference type;
- has a non-static data member or a direct or virtual base class that cannot be copy-assigned (overload resolution for the copy assignment fails, or selects a deleted or inaccessible function);
- is a union-like class, and has a variant member whose corresponding assignment operator is non-trivial.
Trivial copy assignment operator
The copy assignment operator for class is trivial if all of the following is true:
- it is not user-provided (meaning, it is implicitly-defined or defaulted) , , and if it is defaulted, its signature is the same as implicitly-defined(until C++14);
- has no virtual member functions;
- has no virtual base classes;
- the copy assignment operator selected for every direct base of is trivial;
- the copy assignment operator selected for every non-static class type (or array of class type) member of is trivial;
A trivial copy assignment operator makes a copy of the object representation as if by std::memmove. All data types compatible with the C language (POD types) are trivially copy-assignable.
Implicitly-defined copy assignment operator
If the implicitly-declared copy assignment operator is neither deleted nor trivial, it is defined (that is, a function body is generated and compiled) by the compiler if odr-used. For union types, the implicitly-defined copy assignment copies the object representation (as by std::memmove). For non-union class types (class and struct), the operator performs member-wise copy assignment of the object's bases and non-static members, in their initialization order, using built-in assignment for the scalars and copy assignment operator for class types.
The generation of the implicitly-defined copy assignment operator is deprecated(since C++11) if has a user-declared destructor or user-declared copy constructor.
If both copy and move assignment operators are provided, overload resolution selects the move assignment if the argument is an rvalue (either a prvalue such as a nameless temporary or an xvalue such as the result of std::move), and selects the copy assignment if the argument is an lvalue (named object or a function/operator returning lvalue reference). If only the copy assignment is provided, all argument categories select it (as long as it takes its argument by value or as reference to const, since rvalues can bind to const references), which makes copy assignment the fallback for move assignment, when move is unavailable.
It is unspecified whether virtual base class subobjects that are accessible through more than one path in the inheritance lattice, are assigned more than once by the implicitly-defined copy assignment operator (same applies to move assignment).
See assignment operator overloading for additional detail on the expected behavior of a user-defined copy-assignment operator.
Run this code
The following behavior-changing defect reports were applied retroactively to previously published C++ standards.
|DR||Applied to||Behavior as published||Correct behavior|
|CWG 2171||C++14||operator=(X&)=default was non-trivial||made trivial|