Pointer member is not initialized in copy constructor - c++

Question

In my application
#include <iostream>
class TestClassA
{
public:
int* m_ptr;
TestClassA(int a)
{
m_ptr = new int(a);
std::cout << "Constructor. this: " << this << " m_ptr: " << m_ptr << std::endl;
}
TestClassA(const TestClassA& copy)
{
std::cout << "Copy Constructor. copy: " << &copy << " -> this: " << this << std::endl;
std::cout << "Copy Constructor. old this->m_ptr: " << m_ptr << std::endl;
delete m_ptr; // not initialized pointer
m_ptr = new int;
std::cout << "Copy Constructor. new this->m_ptr: " << m_ptr << std::endl;
*m_ptr = *copy.m_ptr;
}
// passing by value, thus a copy constructor calls first
TestClassA& operator=(TestClassA tmp)
{
std::cout << "Copy assignment " << this << " <- " << &tmp << std::endl;
std::swap(m_ptr, tmp.m_ptr);
return *this;
}
~TestClassA()
{
std::cout << "Destructor " << this << std::endl;
delete m_ptr;
m_ptr = nullptr;
}
};
void testAssignment()
{
TestClassA tca1(1);
std::cout << "tca1.m_ptr: " << tca1.m_ptr << std::endl;
TestClassA tca2(2);
std::cout << "tca2.m_ptr: " << tca2.m_ptr << std::endl;
tca2 = tca1;
}
int main()
{
testAssignment();
return 0;
}
When I call assignment operator receiving arguments by value, copy constructor calls. I guess it is to create a temporary variable and to copy the state of tcs1 to it. The issue is that m_ptr member of this temporary is not initialized, so I can't delete previous m_ptr value to write a new one. What is the proper way of implementing copy constructor in this case?

Answer 1

Copy constructor is a constructor, not an assignment operator. The diffetence is precisely the absence of preexisting resources to destroy. You don't need to destroy anything, just initialize.
The copy constructor is called because you did not make it accept a const reference:
TestClassA& operator=(const TestClassA& tmp)
// ^ ^
It is the tmp parameter that is initialized in the example, not the this of the operator.
Of course, you'll need a local variable to get the swap trick to work, but at least it will be explicit in your code.