Copy Constructor Assignment Operator Difference Between Republicans

Copy constructor and assignment operator, are the two ways to initialize one object using another object. The fundamental difference between the copy constructor and assignment operator is that the copy constructor allocates separate memory to both the objects, i.e. the newly created target object and the source object.The assignment operator allocates same memory location to the newly created target object as well as source object.

Let us study the difference between the copy constructor and assignment operator.

Content: Copy Constructor Vs Assignment Operator

  1. Comparison Chart
  2. Definition
  3. Key Differences
  4. Conclusion

Comparison Chart

Basis for ComparisonCopy ConstructorAssignment Operator
BasicThe copy constructor is an overloaded constructor.The assignment operator is a bitwise operator.
MeaningThe copy constructor initializes the new object with an already existing object.The assignment operator assigns the value of one object to another object both of which are already in existence.
Syntaxclass_name(cont class_name &object_name) {
//body of the constructor
}
class_name Ob1, Ob2;
Ob2=Ob1;
Invokes(1)Copy constructor invokes when a new object is initialized with existing one.
(2)The object passed to a function as a non-reference parameter.
(3)The object is returned from the function.
The assignment operator is invoked only when assigning the existing object a new object.
Memory AllocationBoth the target object and the initializing object shares the different memory locations.Both the target object and the initializing object shares same allocated memory.
DefaultIf you do not define any copy constructor in the program, C++ compiler implicitly provides one.If you do not overload the "=" operator, then a bitwise copy will be made.

Definition of Copy Constructor

A “copy constructor” is a form of an overloaded constructor. A copy constructor is only called or invoked for initialization purpose. A copy constructor initializes the newly created object by an another existing object. When a copy constructor is used to initialize the newly created target object, then both the target object and the source object shares different memory location. Changes done to the source object do not reflect in the target  object. The general form of the copy constructor is

class_ name (class_name &object_name){ . // body of copy constructor . } // object_name refers to the object on the right-hand side of the initialization.

If the programmer does not create a copy constructor in a C++ program, then the compiler implicitly provides a copy constructor. An implicit copy constructor provided by the compiler does the member-wise copy of the source object. But, sometimes the member-wise copy is not sufficient, as the object may contain a pointer variable. Copying a pointer variable means, we copy the address stored in the pointer variable, but we do not want to copy address stored in the pointer variable, instead we want to copy what pointer points to. Hence, there is a need of explicit ‘copy constructor’ in the program to solve this kind of problems.

A copy constructor is invoked in three conditions as follow:

  • Copy constructor invokes when a new object is initialized with existing one.
  • The object passed to a function as a non-reference parameter.
  • The object is returned from the function.

Let us understand copy constructor with an example.

class copy{ int num; public: copy(){ } //default constructor copy(int a){ //initializing constructor num=a; } copy( copy &c ){ //Copy constructor num = c.num; } void show( ){ cout<< num; } }; int main(){ copy A(200); //Object A created and initialized copy B(A); // Copy constructor called copy C=A; // Copy constructor called copy D; D=A; //copy constructor not called because object D not newly created object. //it is an assignment operation. return 0; }

In the code above, I had explicitly declared a constructor “copy( copy &c )”. This copy constructor is being called when object B is initialized using object A. Second time it is called when object C is being initialized using object A. When object D is initialized using object A the copy constructor is not called because when D is being initialized it is already in the existence, not the newly created one. Hence, here the assignment operator is invoked.

Definition of Assignment Operator

The assignment operator is an assigning operator of C++.  The “=” operator is used to invoke the assignment operator. It copies the data in one object identically to another object. The assignment operator copies one object to another member-wise. If you do not overload the assignment operator, it performs the bitwise copy. Therefore, you need to overload the assignment operator.

class copy{ int num; public: copy(){ } //default constructor copy(int a){ //initializing constructor num=a; } void show( ){ cout<< num; } }; int main(){ copy A(200); //Object A created and initialized copy B(300); // Object B created and initialized B=A; // assignment operator invoked copy C; C=A; // assignment operator invoked return 0; }

In above code when objectA is assigned to object B the assignment operator is being invoked as both the objects are already in existence. Similarly, same is the case when object C is initialized with object A.

When the bitwise assignment is performed both the object shares the same memory location and changes in one object reflect in another object.

Key Differences Between Copy Constructor and Assignment Operator

  1. A copy constructor is an overloaded contructor where as an assignment operator is a bitwise operator.
  2. Using copy constructor you can initialize a new object with an already existing object. On the other hand, an assignment operator copies one object to the other object, both of which are already in existance.
  3. A copy construcor is initialized whenever a new object is initialized with an already existing object, when an object is passed to a function as a non refrence parameter, or when an object is returned from a function. On the other hand, an assignment operator is invoked only when an object is being assigned to another object.
  4. When an object is being initialized using copy constructor, the initializing object and the initialized object shares the different memory location. On the other hand, when an object is being initialized using an assignment operator then the initialized and initializing objects shares the same memory location.
  5. If you do not explicitly define a copy constructor then the compiler provides one. On the other hand, if you do not overload an assigment operator then a bitwise copy operation is performed.

Conclusion:

The Copy constructor is best for copying one object to another when the object contains raw pointers.

Filed Under: Programming

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
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
constructors:



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
reasons.

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:


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