C++ – What’s the use of the private copy constructor in c++

assignment-operatorccopy-constructor

Why do people define a private copy constructor?

When is making the copy constructor and the assignment operator private a good design?

If there are no members in the class which are pointers or handles to a unique object (like file name), then wat other cases are there where private copy constructor is a good idea?

Same question apply for assignment operator. Given that majority of C++ revolves around copying of objects and passing by reference, are there any good designs which involve private copy constructor?

Best Answer

One use case is the singleton pattern where there can only be exactly one instance of a class. In this case, you need make your constructors and assignment operator= private so that there is no way of creating more than one object. The only way to create an object is via your GetInstance() function as shown below.

// An example of singleton pattern
class CMySingleton
{
public:
  static CMySingleton& GetInstance()
  {
    static CMySingleton singleton;
    return singleton;
  }

// Other non-static member functions
private:
  CMySingleton() {}                                  // Private constructor
  ~CMySingleton() {}
  CMySingleton(const CMySingleton&);                 // Prevent copy-construction
  CMySingleton& operator=(const CMySingleton&);      // Prevent assignment
};

int main(int argc, char* argv[])
{
  // create a single instance of the class
  CMySingleton &object = CMySingleton::GetInstance();

  // compile fail due to private constructor
  CMySingleton object1;
  // compile fail due to private copy constructor
  CMySingleton object2(object);
  // compile fail due to private assignment operator
  object1 = object;

  // ..
  return 0;
}

Related Solutions

C++ – What are the differences between a pointer variable and a reference variable in C++

A pointer can be re-assigned:

int x = 5;
int y = 6;
int *p;
p = &x;
p = &y;
*p = 10;
assert(x == 5);
assert(y == 10);

A reference cannot be re-bound, and must be bound at initialization:

int x = 5;
int y = 6;
int &q; // error
int &r = x;

A pointer variable has its own identity: a distinct, visible memory address that can be taken with the unary & operator and a certain amount of space that can be measured with the sizeof operator. Using those operators on a reference returns a value corresponding to whatever the reference is bound to; the reference’s own address and size are invisible. Since the reference assumes the identity of the original variable in this way, it is convenient to think of a reference as another name for the same variable.
```
int x = 0;
int &r = x;
int *p = &x;
int *p2 = &r;

assert(p == p2); // &x == &r
assert(&p != &p2);
```

You can have arbitrarily nested pointers to pointers offering extra levels of indirection. References only offer one level of indirection.

int x = 0;
int y = 0;
int *p = &x;
int *q = &y;
int **pp = &p;

**pp = 2;
pp = &q; // *pp is now q
**pp = 4;

assert(y == 4);
assert(x == 2);

A pointer can be assigned nullptr, whereas a reference must be bound to an existing object. If you try hard enough, you can bind a reference to nullptr, but this is undefined and will not behave consistently.

/* the code below is undefined; your compiler may optimise it
 * differently, emit warnings, or outright refuse to compile it */

int &r = *static_cast<int *>(nullptr);

// prints "null" under GCC 10
std::cout
    << (&r != nullptr
        ? "not null" : "null")
    << std::endl;

bool f(int &r) { return &r != nullptr; }

// prints "not null" under GCC 10
std::cout
    << (f(*static_cast<int *>(nullptr))
        ? "not null" : "null")
    << std::endl;

You can, however, have a reference to a pointer whose value is nullptr.

Pointers can iterate over an array; you can use ++ to go to the next item that a pointer is pointing to, and + 4 to go to the 5th element. This is no matter what size the object is that the pointer points to.
A pointer needs to be dereferenced with * to access the memory location it points to, whereas a reference can be used directly. A pointer to a class/struct uses -> to access its members whereas a reference uses a ..
References cannot be put into an array, whereas pointers can be (Mentioned by user @litb)
Const references can be bound to temporaries. Pointers cannot (not without some indirection):
```
const int &x = int(12); // legal C++
int *y = &int(12); // illegal to take the address of a temporary.
```
This makes const & more convenient to use in argument lists and so forth.

C++ – What does the explicit keyword mean

The compiler is allowed to make one implicit conversion to resolve the parameters to a function. What this means is that the compiler can use constructors callable with a single parameter to convert from one type to another in order to get the right type for a parameter.

Here's an example class with a constructor that can be used for implicit conversions:

class Foo
{
public:
  // single parameter constructor, can be used as an implicit conversion
  Foo (int foo) : m_foo (foo) 
  {
  }

  int GetFoo () { return m_foo; }

private:
  int m_foo;
};

Here's a simple function that takes a Foo object:

void DoBar (Foo foo)
{
  int i = foo.GetFoo ();
}

and here's where the DoBar function is called:

int main ()
{
  DoBar (42);
}

The argument is not a Foo object, but an int. However, there exists a constructor for Foo that takes an int so this constructor can be used to convert the parameter to the correct type.

The compiler is allowed to do this once for each parameter.

Prefixing the explicit keyword to the constructor prevents the compiler from using that constructor for implicit conversions. Adding it to the above class will create a compiler error at the function call DoBar (42). It is now necessary to call for conversion explicitly with DoBar (Foo (42))

The reason you might want to do this is to avoid accidental construction that can hide bugs.
Contrived example:

You have a MyString class with a constructor that constructs a string of the given size. You have a function print(const MyString&) (as well as an overload print (char *string)), and you call print(3) (when you actually intended to call print("3")). You expect it to print "3", but it prints an empty string of length 3 instead.

Best Answer

Related Solutions

C++ – What are the differences between a pointer variable and a reference variable in C++

C++ – What does the explicit keyword mean

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