Theory of copy constructor

I think you are overdoing this at this point. This is your 5th or 6th thread on the same topic (I have deleted at least 4 duplicates so far).

If there were no copy constructors at all, you would not be able to call a method that takes non-reference arguments. Nor would you ever be able to assign one object to another. Or in short, the capabilities of the programming language would be very seriously limited.

I think this is about as "theoretical" as it gets,
read section 12.8 ISO/IEC JTC1 SC22 WG21 N3092[^]

You might be better off reading:
Thinking in C++ 2nd Edition[^] by Bruce Eckel

It's a free electronic book

Regards
Espen Harlinn

class MyString
{
public:
  MyString(const TCHAR* str){ buff=0; length=0; operator =(str); }

#ifdef USE_COPY_CONSTRUCTOR
  MyString(MyString& copy){ buff=0; length=0; operator = (copy); }
#endif // USE_COPY_CONSTRUCTOR

  ~MyString(){ if(buff) free(buff); }
  
  MyString& operator = (const TCHAR* str){ if(buff) free(buff); buff=_tcsdup(str); len=_tcslen(buff); return *this; }
  operator const TCHAR* () { return buff?buff:__TEXT(""); }

private:
  TCHAR*        buff;
  unsigned int  length;
};


{
  MyString  aaa(__TEXT("aaa"));
  MyString  bbb = aaa;
} // --> crash

#define USE_COPY_CONSTRUCTOR

{
  MyString  aaa(__TEXT("aaa"));
  MyString  bbb = aaa;
} // --> ok

by default members are copied by memcopy.

normally copy const are used when your class contains shareable resources like pointers,handles etc.

In copy constructor of those classes you should create a new instance and pass to the destination object. this is the normal way. it is up to you.

if you are sure about sharing a pointer or handle among objects of your classes and its safe in your application/design, you can live with that.

Hopes it will give you the idea. :)