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.
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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|
Published by jsmith
Jan 27, 2010 (last update: Aug 20, 2010)
Copy constructors, assignment operators, and exception safe assignment
Score: 4.2/5 (2810 votes)
What is a copy constructor?A copy constructor is a special constructor for a class/struct that is
used to make a copy of an existing instance. According to the C++
standard, the copy constructor for MyClass must have one of the
Note that none of the following constructors, despite the fact that
they could do the same thing as a copy constructor, are copy
or my personal favorite way to create an infinite loop in C++:
When do I need to write a copy constructor?First, you should understand that if you do not declare a copy
constructor, the compiler gives you one implicitly. The implicit
copy constructor does a member-wise copy of the source object.
For example, given the class:
the compiler-provided copy constructor is exactly equivalent to:
In many cases, this is sufficient. However, there are certain
circumstances where the member-wise copy version is not good enough.
By far, the most common reason the default copy constructor is not
sufficient is because the object contains raw pointers and you need
to take a "deep" copy of the pointer. That is, you don't want to
copy the pointer itself; rather you want to copy what the pointer
points to. Why do you need to take "deep" copies? This is
typically because the instance owns the pointer; that is, the
instance is responsible for calling delete on the pointer at some
point (probably the destructor). If two objects end up calling
delete on the same non-NULL pointer, heap corruption results.
Rarely you will come across a class that does not contain raw
pointers yet the default copy constructor is not sufficient.
An example of this is when you have a reference-counted object.
boost::shared_ptr<> is example.
Const correctnessWhen passing parameters by reference to functions or constructors, be very
careful about const correctness. Pass by non-const reference ONLY if
the function will modify the parameter and it is the intent to change
the caller's copy of the data, otherwise pass by const reference.
Why is this so important? There is a small clause in the C++ standard
that says that non-const references cannot bind to temporary objects.
A temporary object is an instance of an object that does not have a
variable name. For example:
is a temporary, because we have not given it a variable name. This
is not a temporary:
because the object's name is s.
What is the practical implication of all this? Consider the following:
Many of the STL containers and algorithms require that an object
be copyable. Typically, this means that you need to have the
copy constructor that takes a const reference, for the above
What is an assignment operator?The assignment operator for a class is what allows you to use
= to assign one instance to another. For example:
There are actually several different signatures that an
assignment operator can have:
(1) MyClass& operator=( const MyClass& rhs );
(2) MyClass& operator=( MyClass& rhs );
(3) MyClass& operator=( MyClass rhs );
(4) const MyClass& operator=( const MyClass& rhs );
(5) const MyClass& operator=( MyClass& rhs );
(6) const MyClass& operator=( MyClass rhs );
(7) MyClass operator=( const MyClass& rhs );
(8) MyClass operator=( MyClass& rhs );
(9) MyClass operator=( MyClass rhs );
These signatures permute both the return type and the parameter
type. While the return type may not be too important, choice
of the parameter type is critical.
(2), (5), and (8) pass the right-hand side by non-const reference,
and is not recommended. The problem with these signatures is that
the following code would not compile:
This is because the right-hand side of this assignment expression is
a temporary (un-named) object, and the C++ standard forbids the compiler
to pass a temporary object through a non-const reference parameter.
This leaves us with passing the right-hand side either by value or
by const reference. Although it would seem that passing by const
reference is more efficient than passing by value, we will see later
that for reasons of exception safety, making a temporary copy of the
source object is unavoidable, and therefore passing by value allows
us to write fewer lines of code.
When do I need to write an assignment operator?First, you should understand that if you do not declare an
assignment operator, the compiler gives you one implicitly. The
implicit assignment operator does member-wise assignment of
each data member from the source object. For example, using
the class above, the compiler-provided assignment operator is
exactly equivalent to:
In general, any time you need to write your own custom copy
constructor, you also need to write a custom assignment operator.
What is meant by Exception Safe code?A little interlude to talk about exception safety, because programmers
often misunderstand exception handling to be exception safety.
A function which modifies some "global" state (for example, a reference
parameter, or a member function that modifies the data members of its
instance) is said to be exception safe if it leaves the global state
well-defined in the event of an exception that is thrown at any point
during the function.
What does this really mean? Well, let's take a rather contrived
(and trite) example. This class wraps an array of some user-specified
type. It has two data members: a pointer to the array and a number of
elements in the array.
Now, assignment of one MyArray to another is easy, right?
Well, not so fast. The problem is, the line
could throw an exception. This line invokes operator= for type T, which
could be some user-defined type whose assignment operator might throw an
exception, perhaps an out-of-memory (std::bad_alloc) exception or some
other exception that the programmer of the user-defined type created.
What would happen if it did throw, say on copying the 3rd element of 10
total? Well, the stack is unwound until an appropriate handler is found.
Meanwhile, what is the state of our object? Well, we've reallocated our
array to hold 10 T's, but we've copied only 2 of them successfully. The
third one failed midway, and the remaining seven were never even attempted
to be copied. Furthermore, we haven't even changed numElements, so whatever
it held before, it still holds. Clearly this instance will lie about the
number of elements it contains if we call count() at this point.
But clearly it was never the intent of MyArray's programmer to have count()
give a wrong answer. Worse yet, there could be other member functions that
rely more heavily (even to the point of crashing) on numElements being correct.
Yikes -- this instance is clearly a timebomb waiting to go off.
This implementation of operator= is not exception safe: if an exception is
thrown during execution of the function, there is no telling what the state
of the object is; we can only assume that it is in such a bad state (ie,
it violates some of its own invariants) as to be unusable. If the object is
in a bad state, it might not even be possible to destroy the object without
crashing the program or causing MyArray to perhaps throw another exception.
And we know that the compiler runs destructors while unwinding the stack to
search for a handler. If an exception is thrown while unwinding the stack,
the program necessarily and unstoppably terminates.
How do I write an exception safe assignment operator?The recommended way to write an exception safe assignment operator is via
the copy-swap idiom. What is the copy-swap idiom? Simply put, it is a two-
step algorithm: first make a copy, then swap with the copy. Here is our
exception safe version of operator=:
Here's where the difference between exception handling and exception safety
is important: we haven't prevented an exception from occurring; indeed,
the copy construction of tmp from rhs may throw since it will copy T's.
But, if the copy construction does throw, notice how the state of *this
has not changed, meaning that in the face of an exception, we can guarantee
that *this is still coherent, and furthermore, we can even say that it is
But, you say, what about std::swap? Could it not throw? Yes and no. The
default std::swap<>, defined in <algorithm> can throw, since std::swap<>
looks like this:
The first line runs the copy constructor of T, which can throw; the
remaining lines are assignment operators which can also throw.
HOWEVER, if you have a type T for which the default std::swap() may result
in either T's copy constructor or assignment operator throwing, you are
politely required to provide a swap() overload for your type that does not
throw. [Since swap() cannot return failure, and you are not allowed to throw,
your swap() overload must always succeed.] By requiring that swap does not
throw, the above operator= is thus exception safe: either the object is
completely copied successfully, or the left-hand side is left unchanged.
Now you'll notice that our implementation of operator= makes a temporary
copy as its first line of code. Since we have to make a copy, we might as
well let the compiler do that for us automatically, so we can change the
signature of the function to take the right-hand side by value (ie, a copy)
rather than by reference, and this allows us to eliminate one line of code: