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++. A trivial copy constructor is a constructor that creates a bytewise copy of the object representation of the argument, and performs no other action. Templates and copy constructor. C / C++ Forums on Bytes. Matthias Spiller wrote: This is a specialization. Do you have a class definition for the. I read that template copy-con. Can someone explain me the whole reason behind putting this restriction and also how to write a copy constructor of template class. Is not a call to the copy constructor; it is a call to operator() [the function call operator] with a parameter of v1. You need to do as R0mai typed. Yeah, i have those in my solution, but did not copy the wole class and adapted it here. I'm using VS2008, could it be compiler relative? thx. Copy constructors, assignment operators, - C++ Articles. What is a copy constructor? A copy constructor is a special constructor for a class/struct that is. According to the C++. My. Class must have one of the. My. Class( const My. Class& other ). My. Class( My. Class& other ). My. Class( volatileconst My. Class& other ). My. Class( volatile My. Class& other ); Note that none of the following constructors, despite the fact thatthey could do the same thing as a copy constructor, are copyconstructors: 1. My. Class( My. Class* other ). Class Template Copy Constructor InheritanceMy. Class( const My. Class* other ); or my personal favorite way to create an infinite loop in C++: My. Class( My. Class other ); When do I need to write a copy constructor? First, you should understand that if you do not declare a copy. The implicit. copy constructor does a member- wise copy of the source object. For example, given the class: 1. My. Class {. std: :string s. My. Class: :My. Class( const My. Class& other ) . In many cases, this is sufficient. However, there are certaincircumstances where the member- wise copy version is not good enough. By far, the most common reason the default copy constructor is notsufficient is because the object contains raw pointers and you needto 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 pointerpoints to. Why do you need to take "deep" copies? This is typically because the instance owns the pointer; that is, theinstance is responsible for calling delete on the pointer at somepoint (probably the destructor). If two objects end up callingdelete on the same non- NULL pointer, heap corruption results. Rarely you will come across a class that does not contain rawpointers yet the default copy constructor is not sufficient. An example of this is when you have a reference- counted object. Const correctness. When passing parameters by reference to functions or constructors, be very. Pass by non- const reference ONLY if. Why is this so important? There is a small clause in the C++ standardthat says that non- const references cannot bind to temporary objects. A temporary object is an instance of an object that does not have avariable name. For example: std: :string( "Hello world" ); is a temporary, because we have not given it a variable name. Thisis not a temporary: std: :string s( "Hello world" ); because the object's name is s. What is the practical implication of all this? Consider the following: 1. Improperly declared function: parameter should be const reference: void print_me_bad( std: :string& s ) {. Properly declared function: function has no intent to modify s: void print_me_good( const std: :string& s ) {. Hello" ). print_me_bad( hello ); // Compiles ok; hello is not a temporary. World" ) ); // Compile error; temporary object. Compile error; compiler wants to construct temporary// std: :string from const char*. Compiles ok. print_me_good( std: :string( "World" ) ); // Compiles ok. Compiles ok Many of the STL containers and algorithms require that an objectbe copyable. Typically, this means that you need to have thecopy constructor that takes a const reference, for the abovereasons. What is an assignment operator? The assignment operator for a class is what allows you to use. For example: 1. 2 My. Class c. 1, c. 2. There are actually several different signatures that anassignment operator can have: (1) My. Class& operator=( const My. Class& rhs ); (2) My. Class& operator=( My. Class& rhs ); (3) My. Class& operator=( My. Class rhs ); (4) const My. Class& operator=( const My. Class& rhs ); (5) const My. Class& operator=( My. Class& rhs ); (6) const My. Class& operator=( My. Class rhs ); (7) My. Class operator=( const My. Class& rhs ); (8) My. Class operator=( My. Class& rhs ); (9) My. Class operator=( My. Class rhs ); These signatures permute both the return type and the parametertype. While the return type may not be too important, choiceof 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 thatthe following code would not compile: 1. My. Class c. 1. c. My. Class( 5, 'a', "Hello World" ); // assuming this constructor exists This is because the right- hand side of this assignment expression isa temporary (un- named) object, and the C++ standard forbids the compilerto pass a temporary object through a non- const reference parameter. This leaves us with passing the right- hand side either by value orby const reference. Although it would seem that passing by constreference is more efficient than passing by value, we will see laterthat for reasons of exception safety, making a temporary copy of thesource object is unavoidable, and therefore passing by value allowsus 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. The. implicit assignment operator does member- wise assignment of. For example, using. My. Class& My. Class: :operator=( const My. Class& other ) {. 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. A function which modifies some "global" state (for example, a referenceparameter, or a member function that modifies the data members of its instance) is said to be exception safe if it leaves the global statewell- defined in the event of an exception that is thrown at any pointduring 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- specifiedtype. It has two data members: a pointer to the array and a number of elements in the array. T >. class My. Array {. Elements. T* p. Elements. size_t count() const { return num. Elements; }. My. Array& operator=( const My. Array& rhs ). Now, assignment of one My. Array to another is easy, right? My. Array< T> :: operator=( const My. Array& rhs ) {. Elements. p. Elements = new T[ rhs. Elements ]. for( size_t i = 0; i < rhs. Elements; ++i ). p. Elements[ i ] = rhs. Elements[ i ]. num. Elements = rhs. num. Elements. return *this. Well, not so fast. The problem is, the line p. Elements[ i ] = rhs. Elements[ i ]; could throw an exception. This line invokes operator= for type T, whichcould be some user- defined type whose assignment operator might throw anexception, perhaps an out- of- memory (std: :bad_alloc) exception or someother exception that the programmer of the user- defined type created. What would happen if it did throw, say on copying the 3rd element of 1. Well, the stack is unwound until an appropriate handler is found. Meanwhile, what is the state of our object? Well, we've reallocated ourarray to hold 1. T's, but we've copied only 2 of them successfully. Thethird one failed midway, and the remaining seven were never even attemptedto be copied. Furthermore, we haven't even changed num. Elements, so whateverit held before, it still holds. Clearly this instance will lie about thenumber of elements it contains if we call count() at this point. But clearly it was never the intent of My. Array's programmer to have count()give a wrong answer. Worse yet, there could be other member functions thatrely more heavily (even to the point of crashing) on num. Elements being correct. Yikes - - this instance is clearly a timebomb waiting to go off. This implementation of operator= is not exception safe: if an exception isthrown during execution of the function, there is no telling what the stateof 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 withoutcrashing the program or causing My. Array 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. What is the copy- swap idiom? Simply put, it is a two- . Here is our. exception safe version of operator=: 1. My. Array< T> :: operator=( const My. Array& rhs ) {. First, make a copy of the right- hand side. My. Array tmp( rhs ). Now, swap the data members with the temporary. Elements, tmp. num. Elements ). std: :swap( p. Elements, tmp. p. Elements ). return *this. Here's where the difference between exception handling and exception safetyis 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 *thishas not changed, meaning that in the face of an exception, we can guaranteethat *this is still coherent, and furthermore, we can even say that it isleft unchanged. But, you say, what about std: :swap? Could it not throw? Yes and no. Thedefault std: :swap< > , defined in < algorithm> can throw, since std: :swap< > looks like this: 1. T >. swap( T& one, T& two ). T tmp( one ). one = two. 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 resultin either T's copy constructor or assignment operator throwing, you are politely required to provide a swap() overload for your type that does notthrow. 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.
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