The definition for a converting constructor is different between C++03 and C++11. In both cases it must be a non-explicit constructor (otherwise it wouldn’t be involved in implicit conversions), but for C++03 it must also be callable with a single argument. That is: struct foo { foo(int x); // 1 foo(char* s, int x = …
From [class.copy]/7, emphasis mine: If the class definition does not explicitly declare a copy constructor, a non-explicit one is declared implicitly. If the class definition declares a move constructor or move assignment operator, the implicitly declared copy constructor is defined as deleted; otherwise, it is defined as defaulted (8.4). The latter case is deprecated if …
No, they are different operators. The copy constructor is for creating a new object. It copies an existing object to a newly constructed object.The copy constructor is used to initialize a new instance from an old instance. It is not necessarily called when passing variables by value into functions or as return values out of …
For building containers you obviously want to use one of the standard containers (such as a std::vector). But this is a perfect example of the things you need to consider when your object contains RAW pointers. If your object has a RAW pointer then you need to remember the rule of 3 (now the rule …
Java does. They’re just not called implicitly like they are in C++ and I suspect that’s your real question. Firstly, a copy constructor is nothing more than: public class Blah { private int foo; public Blah() { } // public no-args constructor public Blah(Blah b) { foo = b.foo; } // copy constructor } Now …
class Foo { private: Foo(); Foo( const Foo& ); // non construction-copyable Foo& operator=( const Foo& ); // non copyable public: static Foo* create(); } If you’re using boost, you can also inherit from noncopyable : http://www.boost.org/doc/libs/1_41_0/boost/noncopyable.hpp EDIT: C++11 version if you have a compiler supporting this feature: class Foo { private: Foo(); public: Foo( …
A bit of clarification as to why it’s preferable to return by reference for operator= versus return by value — as the chain a = b = c will work fine if a value is returned. If you return a reference, minimal work is done. The values from one object are copied to another object. …
The copy constructor generated by the compiler does member-wise copying. Sometimes that is not sufficient. For example: class Class { public: Class( const char* str ); ~Class(); private: char* stored; }; Class::Class( const char* str ) { stored = new char[srtlen( str ) + 1 ]; strcpy( stored, str ); } Class::~Class() { delete[] stored; …
Yes. There are two common options. One – which is generally discouraged – is to call the operator= from the copy constructor explicitly: MyClass(const MyClass& other) { operator=(other); } However, providing a good operator= is a challenge when it comes to dealing with the old state and issues arising from self assignment. Also, all members …
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, …