Published by jsmith
Jan 27, 2010 (last update: Aug 20, 2010)
Copy constructors, assignment operators, and exception safe assignment
Score: 4.2/5 (2813 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:
In the C++programming language, a copy constructor is a special constructor for creating a new objectas a copy of an existing object. Copy constructors are the standard way of copying objects in C++, as opposed to cloning, and have C++-specific nuances.
The first argument of such a constructor is a reference to an object of the same type as is being constructed (const or non-const), which might be followed by parameters of any type (all having default values).
Normally the compiler automatically creates a copy constructor for each class (known as an implicit copy constructor) but for special cases the programmer creates the copy constructor, known as a user-defined copy constructor. In such cases, the compiler does not create one. Hence, there is always one copy constructor that is either defined by the user or by the system.
A user-defined copy constructor is generally needed when an object owns pointers or non-shareable references, such as to a file, in which case a destructor and an assignment operator should also be written (see Rule of three).
Copying of objects is achieved by the use of a copy constructor and an assignment operator. A copy constructor has as its first parameter a (possibly const or volatile) reference to its own class type. It can have more arguments, but the rest must have default values associated with them. The following would be valid copy constructors for class :
The first one should be used unless there is a good reason to use one of the others. One of the differences between the first and the second is that temporaries can be copied with the first. For example:
Another difference between them is the obvious:
The form of the copy constructor is used when it is necessary to modify the copied object. This is very rare but it can be seen used in the standard library's . A reference must be provided:
The following are invalid copy constructors (Reason - is not passed as reference) :
because the call to those constructors would require a copy as well, which would result in an infinitely recursive call.
The following cases may result in a call to a copy constructor:
- When an object is returned by value
- When an object is passed (to a function) by value as an argument
- When an object is thrown
- When an object is caught
- When an object is placed in a brace-enclosed initializer list
These cases are collectively called copy-initialization and are equivalent to:
It is however, not guaranteed that a copy constructor will be called in these cases, because the C++ Standard allows the compiler to optimize the copy away in certain cases, one example being the return value optimization (sometimes referred to as RVO).
An object can be assigned value using one of the two techniques:
- Explicit assignment in an expression
Explicit assignment in an expression
An object can be initialized by any one of the following ways.
a. Through declaration
b. Through function arguments
c. Through function return value
The copy constructor is used only for initializations, and does not apply to assignments where the assignment operator is used instead.
The implicit copy constructor of a class calls base copy constructors and copies its members by means appropriate to their type. If it is a class type, the copy constructor is called. If it is a scalar type, the built-in assignment operator is used. Finally, if it is an array, each element is copied in the manner appropriate to its type.
By using a user-defined copy constructor the programmer can define the behavior to be performed when an object is copied.
These examples illustrate how copy constructors work and why they are required sometimes.
Implicit copy constructor
Let us consider the following example:
Output10 15 10 23 15 10
As expected, timmy has been copied to the new object, timmy_clone. While timmy's age was changed, timmy_clone's age remained the same. This is because they are totally different objects.
The compiler has generated a copy constructor for us, and it could be written like this:
So, when do we really need a user-defined copy constructor? The next section will explore that question.
User-defined copy constructor
Now, consider a very simple dynamic array class like the following:
Output25 25 Segmentation fault
Since we did not specify a copy constructor, the compiler generated one for us. The generated constructor would look something like:
The problem with this constructor is that it performs a shallow copy of the data pointer. It only copies the address of the original data member; this means they both share a pointer to the same chunk of memory, which is not what we want. When the program reaches line (1), copy's destructor gets called (because objects on the stack are destroyed automatically when their scope ends). Array's destructor deletes the data array of the original, therefore when it deleted copy's data, because they share the same pointer, it also deleted first's data. Line (2) now accesses invalid data and writes to it! This produces the infamous segmentation fault.
If we write our own copy constructor that performs a deep copy then this problem goes away.
Here, we are creating a new int array and copying the contents to it. Now, other's destructor deletes only its data, and not first's data. Line (2) will not produce a segmentation fault anymore.
Instead of doing a deep copy right away, there are some optimization strategies that can be used. These allow you to safely share the same data between several objects, thus saving space. The copy-on-write strategy makes a copy of the data only when it is written to. Reference counting keeps the count of how many objects are referencing the data, and will delete it only when this count reaches zero (e.g. boost::shared_ptr).
Copy constructors and templates
Contrary to expectations, a template copy constructor is not a user-defined copy constructor. Thus it is not enough to just have:
(Note that the type of can be .) A user-defined, non-template copy constructor must also be provided for construction of Array from Array.
Bitwise copy constructor
There is no such thing as "bitwise copy constructor" in C++. However, the default generated copy constructor copies by invoking copy constructors on members, and for a raw pointer member this will copy the raw pointer (i.e. not a deep copy).
Logical copy constructor
A logical copy constructor makes a true copy of the structure as well as its dynamic structures. Logical copy constructors come into the picture mainly when there are pointers or complex objects within the object being copied.
Explicit copy constructor
An explicit copy constructor is one that is declared explicit by using the explicit keyword. For example:
It is used to prevent copying of objects at function calls or with the copy-initialization syntax.