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Answered 2009-07-19 05:08:01

A copy constructor usually refers to a constructor which takes an object, and returns a copy of that object. I can think of no way to overload the constructor without changing its functionality.

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Constructor overloading, just like any function's overloading, is where more than one configuration of parameters exists for the function. Based on the number and type of the parameters, different versions of the function can be resolved by the linker. This is typically used in the constructor as the default constructor (no parameters), the copy constructor (one reference parameter of the same type as the class), and the conversion constructor (any other combination of parameters).

A copy constructor is a type of constructor which constructs the object of the class from another object of the same class. The copy constructor accepts a reference to its own class as a parameter. A valid copy constructor can be specified as, class A { ...... ...... public: A(&A); }

No. Java does not support copy constructor

Overloading a function simply means providing the same function name with different argument types. Class constructors are no different. In fact, even if you declare no constructors in a class, there will be two compiler-generated constructor overloads provided for you: a default constructor; and a copy constructor. If you declare any other constructors, the compiler-generated default constructor will no longer be generated. You must declare your own default constructor if you require one. The copy constructor is always generated, however the default implementation only performs a member-wise copy of the class members. If your class contains a pointer to allocated memory you must provide your own copy constructor to perform a deep-copy of those pointers, so each instances "owns" its own copy of the memory.

What is the advantage of user-defined copy constructor

You overload a constructor by declaring two or more constructors in a class, each with different signatures. When no constructor is specified, a default constructor and a copy constructor are implied. Both can be overridden. The default constructor can also be overloaded if all arguments are given default values in the declaration. You can add as many construction overloads as required in order to initialise your class. If your class has many members to initialise, it may be helpful to use a structure and an overloaded constructor to accept the structure. For every constructor that has exactly one argument, there should also be an equivalent assignment operator overload. The copy constructor assignment overload is implied if not specified.

It is meaningless. Copy constructors cannot be overloaded. You either use the compiler-generated default copy constructor or you define your own. Either way, there can only ever be one copy constructor. The purpose of the copy constructor is to construct a new instance of a class (a new object) from an existing instance of the same class (an existing object). By default, the new object's members will be a bitwise copy (a shallow copy) of the existing object's members. If the class acquires a resource through a member pointer, a user-defined copy constructor must be provided in order to perform a deep copy of that pointer, otherwise you end up with two objects sharing the same resource. This problem does not exist when using smart pointers or resource handles rather than raw pointers.

Because if it's not by reference, it's by value. To do that you make a copy, and to do that you call the copy constructor. But to do that, we need to make a new value, so we call the copy constructor, and so on... (You would have infinite recursion because "to make a copy, you need to make a copy".)

Without a copy constructor the only way to copy an object would be to instantiate a new object (or use an existing object) and then assign another object's value to it. However, it would be very odd indeed to have a copy assignment operator without a matching copy constructor. If you have one, you must have both. If you do not need to copy an object of a particular class, however, you can simply delete both the copy constructor and the copy assigment operator for that class. Any attempt to copy or copy assign would then result in a compile-time error.

class X { public: X(); // default constructor X(const X& x); // copy constructor // ... }; int main(void) { X objx1; //normal ctor X objx2 = x1; // copy ctor X x3(x2); // copy ctor }

Java, unlike C++ does not support copy constructors.

The meaning of copy constructor is a special programmer of the C++ language that copys existing coding projects and enhances or makes them better projects.

More or less as you would any other function, except there is no return type (not even void) and an initialisation list can be placed between the declaration and the definition that follows it. The initialisation list allows your constructor to call base class constructors besides the default constructor as well as initialise member variables according to the parameters passed to your constructor. The constructor's name must be the name of the class. Note that if you don't declare any constructors, the compiler generates both a default and copy constructor. If any constructor is declared you lose the default constructor unless you declare one yourself. The copy constructor is always present but must be overridden if your class contains pointers to memory allocated to the class itself. If you don't, the compiler generated copy constructor will perform a member-wise copy of the member variables, resulting in a shallow copy of the pointers themselves, rather than a deep copy of the memory they point to. The copy constructor must accept a constant reference to the same class of object.

Assignment(=) operator is a special operator that will be provided by the constructor to the class when programmer has not provided(overloaded) as member of the class.(like copy constructor). When programmer is overloading = operator using friend function, two = operations will exists: 1) compiler is providing = operator 2) programmer is providing(overloading) = operator by friend function. Then simply ambiguity will be created and compiler will gives error. Its compilation error.

Whenever a new object is instantiated, the appropriate class constructor for that object is invoked automatically. At compile time, argument-dependent lookup is used to determined which specific constructor to invoke. When passing objects to functions by value, the class copy constructor is invoked. If there is no copy constructor for the class, objects of this class cannot be passed by value. When returning objects from functions by value, the class move constructor is invoked. If there is no move constructor for the class, the class copy constructor is invoked instead. If neither exists, objects of this class cannot be returned by value.

Constructors are not a requirement of CPP. A default constructor and copy constructor are automatically generated by the compiler for every class you create when no constructors are declared. The only time you need to declare a constructor is when you wish to override the default behaviour of the generated constructors, to ensure the class is correctly initialised. When any constructor is declared, the default constructor is no longer generated by the compiler -- you must define your own default constructor (one that has no parameters, or where all the parameters have default values). The copy constructor is always generated for you regardless of how many other constructors you declare. But if the class contains pointers to allocated memory that is "owned" by the class then you must override the generated copy constructor with your own copy constructor. This is to ensure the memory is deep copied (the generated copy constructor only performs a shallow, member-wise copy of the members). Otherwise two objects of the same class will end up pointing at the same memory, which would be disastrous if either one were to be deleted. The other instance would be automatically invalidated because it would point to memory that was released by the other instance's destructor.

No. The parameter of the copy constructor is always passed by reference, specifically, a const reference to the class.Think about it. If it were passed by value, then the compiler would already know how to copy the object into the formal parameter, but the purpose of the copy constructor is to provide the code to copy the object, so its kind of a "cart before the horse" thing to think about call by value here.

Default constructor: X()Copy constructor: X(const X&)Copy assignment operator: X& operator=(const X&)Move constructor: X(X&&)Move assignment operator: X& operator=(XX&)Destructor: ~X()By default, the compiler will generate each of these operations if a program uses it. However, if the programmer declares any constructor for a class, the default constructor for that class is not generated. If the programmer declares a copy or move operation, no copy, move or destructor is generated. If the programmer declares a destructor, no move operation is generated (a copy constructor is generated for backward compatibility).We can also suppress generation of specific operations with the =delete pseudo-initialiser:class X {public:X (const X&) =delete; // suppress the compiler-generated copy operationsX& operator=(const X&) =delete;// ...};

There is no specific keyword for a constructor in C++. Simply define and declare a method of the class with the same name as the class and it will be a constructor. A constructor with no arguments is the default constructor, a constructor with one argument of class type is the copy constructor, and a constructor with one argument of some other type is the conversion constructor. You can provide other overloaded constructors if you want.

The purpose of constructor in object oriented programming is to initialize data. Likewise copy constructors are used to initialize an object with data from another object.

The copy constructor is a special type of constructor. It is identified by the fact that it always accepts a reference to an instance of the same class via its first argument. If there are any additional arguments in the copy constructor, they must have default values. If default values are not provided, the constructor cannot be regarded as a copy constructor. The purpose of the copy constructor is to make an exact copy of an existing object. The reason we must pass the existing object by reference is simply because there is no other way to do it. The compiler won't allow you to pass the exiting object by value because whenever you pass an object by value you automatically invoke the object's copy constructor. Thus the copy constructor would recursively invoke itself. The only other option would be to pass the instance by pointer, but since pointers may be NULL, this is less efficient than passing by reference. References can never be NULL; they must always refer to an existing instance. A NULL reference would result in an invalid program. So although it is possible to construct a copy from a pointer, the extra checks required to ensure the pointer is non-NULL make it less efficient. The following example should demonstrate the difference more clearly: #include<iostream> class foo { int m_data; public: foo(): m_data(0) {} // default constructor foo(const foo& f): m_data(f.m_data) {} // copy constructor foo(const foo* f): m_data(f?f->m_data:0) {} // alternate constructor }; int main() { foo a; // invokes the default constructor foo b(a); // invokes the copy constructor foo c(&b); // invokes the alternate constructor } Since the purpose of the copy constructor is to make a copy of an existing object without altering the existing object, it makes sense to pass a const reference (as shown above), thus assuring the caller that the existing instance's immutable members will not change. Note also that when passing instances of foo by value, the copy constructor is always invoked, never the alternate constructor. As far as the compiler is concerned, the alternate constructor is just another constructor like any other. But since it serves no purpose that isn't already fulfilled by the copy constructor, there's no point in declaring it; it is surplus to requirements. Note also that even if you do not declare your own copy constructor, the compiler will automaticallly generate one for you. However, the compiler-generated copy constructor will perform a member-wise copy (a shallow copy). In many cases this is adequate. But if your class encapsulates unshared memory then you must provide your own copy constructor to ensure that that memory is deep copied. If not, both instances will end up sharing the same memory. This will prove disasterous when one instance is deleted, because the remaining instance(s) would end up pointing at memory that is no longer valid. By deep copying the memory, you ensure that each instance points to its own copy of the memory.

There is no such thing as an explicit copy constructor. Explicit constructors are required when we wish to prevent narrowing through implicit conversion of argument values, but a copy constructor's argument is a constant reference to an existing object of the same type, thus there is no implicit conversion and therefore no narrowing. That is, copying an object of type T or of a type derived from type T will always create a new object of type T. If we want to copy a type derived from a T, then we need to invoke that derived type's copy constructor, not the copy constructor for a T.

If you do not declare a copy constructor, the compiler will generate one for you. However the generated copy constructor only performs a shallow-copy (member-wise copy). This is fine for most values, but if the class contains pointers to allocated memory that would result in two separate classes pointing at the same memory. If the memory was allocated outside of the class (is not owned by the class), this is generally OK. But if the memory is owned by one of the classes, destroying either class would invalidate the other. Therefore you must declare a copy constructor and perform a deep-copy of the memory so that each copy owns its own data.

A copy constructor gets called any time an object needs to be copied. Unlike in some of the newer languages like Java, you can chose to pass objects either by reference or by value. When you pass by reference, only the address of the function is copied. However, if you pass by value, the whole object must be copied. In order to copy the object, the copy constructor will get called. If the copy constructor's parameter is not a reference, then the object will get passed by value. When the object gets passed by value, it needs to get copied so it will call the same copy constructor. Because the object is still being passed by value it'll call itself again to create a copy of itself. The copy constructor will continue to call itself until the stack overflows.

Passing objects by value automatically creates a temporary copy of the object on the call stack. So even if it were possible to pass by value in a copy constructor, the call would not only allocate memory for the copy, it would call the copy constructor to initialise that memory, causing the copy constructor to iterate over and over until there was no more memory left in the call stack. Ultimately, your program crashes without ever having made a copy. Note that a copy constructor not only accepts objects by reference but by constant reference. This is because the copy constructor is only interested in copying the object's members, not in changing them. As a general rule, if an object can be passed to any function by constant reference then that is the way to go. If it is a non-constant reference, then the assumption is the object's immutable members will be altered by the function in some way. If you do not want any changes reflected in your object, then you must copy the object first, which is essentially what happens when you pass the object by value.

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