Copy Constructor And Assignment Operator Example

C++ gives almost god-like powers to the designer of a class. Object "life cycle" management means taking complete control over the behavior of objects during birth, reproduction, and death. You have already seen how constructors manage the birth of an object and how destructors are used to manage the death of an object. This section investigates the reproduction process: the use of copy constructors and assignment operators.

A copy constructor is a constructor that has a prototype like this:

ClassName(const ClassName & x);

The purpose of a copy constructor is to create an object that is an exact copy of an existing object of the same class.

An assignment operator for a class overloads the symbol and gives it a meaning that is specific to the class. There is one particular version of the assignment operator that has the following prototype:

ClassName& operator=(const ClassName& x);

Because it is possible to have several different overloaded versions of the in a class, we call this particular version the copy assignment operator.

Example 2.16. src/lifecycle/copyassign/fraction.h

[ . . . . ] class Fraction { public: Fraction(int n, int d) ; Fraction(const Fraction& other) ; Fraction& operator=(const Fraction& other) ; Fraction multiply(Fraction f2) ; static report() ; private: int m_Numer, m_Denom; static int s_assigns; static int s_copies; static int s_ctors; }; [ . . . . ]

Regular constructor

Copy constructor

Copy assignment operator

The version of in Example 2.16 has three counters, defined in Example 2.17, so that you can count the total number of times each member function is called. This should help you to better understand when objects are copied.

Example 2.17. src/lifecycle/copyassign/fraction.cpp

[ . . . . ] int Fraction::s_assigns = 0; int Fraction::s_copies = 0; int Fraction::s_ctors = 0; Fraction::Fraction(const Fraction& other) : m_Numer(other.m_Numer), m_Denom(other.m_Denom) { ++s_copies; } Fraction& Fraction::operator=(const Fraction& other) { if (this != &other) { m_Numer = other.m_Numer; m_Denom = other.m_Denom; ++s_assigns; } return *this; } [ . . . . ]

Static member definitions.

operator=() should always do nothing in the case of self- assignment.

operator=() should always return *this, to allow for chaining i.e. a=b=c.

Example 2.18 uses this class to create, copy, and assign some objects.

Example 2.18. src/lifecycle/copyassign/copyassign.cpp

#include <> #include "fraction.h" int main() { cout(stdout); Fraction twothirds(2,3); Fraction threequarters(3,4); Fraction acopy(twothirds); Fraction f4 = threequarters; cout << "after declarations - " << Fraction::report(); f4 = twothirds; cout << "\nbefore multiply - " << Fraction::report(); f4 = twothirds.multiply(threequarters); cout << "\nafter multiply - " << Fraction::report() << endl; return 0; }

Using 2-arg constructor.

Using copy constructor.

Also using copy constructor.


Several objects are created here.

Here is the output of this program.

copyassign> ./copyassign after declarations - [assigns: 0 copies: 2 ctors: 2] before multiply - [assigns: 1 copies: 2 ctors: 2] after multiply - [assigns: 2 copies: 3 ctors: 3] copyassign>

As you can see, the call to creates three objects. Can you explain why?

2.11. Copy Constructors and Assignment Operators

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
following signatures:

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 correctness

When 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
left unchanged.

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:

One thought on “Copy Constructor And Assignment Operator Example

Leave a Reply

Your email address will not be published. Required fields are marked *