Copy constructor with pointers

Question

I have recently discovered that when I have pointers within a class, I need to specify a Copy constructor.

To learn that, I have made the following simple code. It c

Answer 1

if it has a pointer to a regular type then

A::A(const A& a):
  pointer_( new int( *a.pointer_ ) )
{
}

if it has a pointer to some base class then

A::A(const &a ):
  pointer_( a.pointer_->clone() )
{
}

Clone is a implementation of a prototype pattern

Don't forget to delete the pointer in the destructor

A::~A()
{
    delete pointer_;
}

To fix your example

TRY::TRY(TRY const & copyTRY){
    int a = *copyTRY.pointer;
    pointer = new int(a);
}

Answer 2

With the statement int* pointer you have just defined a pointer but has not allocated any memory. First you should make it point to a proper memory location by allocating some memory like this: int* pointer = new int. Then in the copy constructor again you have to allocate the memory for the copied object. Also, don't forget to release the memory using delete in the destructor.

I hope this example helps:

class B
{

public:
    B();
    B(const B& b);
    ~B();
    void setVal(int val);

private:
    int* m_p;
};

B::B() 
{
    //Allocate the memory to hold an int
    m_p = new int;

    *m_p = 0;
}

B::B(const B& b)
{
    //Allocate the memory first
    m_p = new int;

    //Then copy the value from the passed object
    *m_p = *b.m_p;
}

B::~B()
{

    //Release the memory allocated
    delete m_p;
    m_p = NULL;
}

void B::setVal(int val)
{
    *m_p = val;
}

Answer 3

When writing a Copy Constructor, you should allocate memory for all members. In your case:

TRY::TRY(TRY const & copyTRY){
    pointer = new int(*(copyTry.pointer));
}

Operator= is somehow similar, but with no memory allocation.

TRY& operator=(TRY const& otherTRY){
      this->a  = *(otherTry.pointer)
      return *this
}