Copy constructors - c++

Question

Can I write a copy constructor by just passing in a pointer instead of the const reference? (Would it be ok if I make sure that I am not going to change any values though?)

Like so:

SampleClass::SampleClass(SampleClass* p)
{
 //do  the necessary copy functionality
}

instead of:

SampleClass::SampleClass(const SampleClass& copyObj)
{
//do the necessary copy
}

Thanks in advance.

---

Thanks everyone. So, if I write a constructor that takes in a pointer( and thought that's my copy constructor), the compiler would still supply with the default copy constructor in which case my constructor( which i thought was my copy constructor) would not be called and the default copy constructor would be called. Got it.

Answer 1

Yes, you can write a constructor that takes a pointer to the object. However, it cannot be called a copy constructor. The very definition of a copy constructor requires you to pass an object of the same class. If you are passing anything else, yes, it's a constructor alright, but not a copy constructor.

Answer 2

You can write a constructor that takes a pointer as an argument.
But the copy constructor is the name we give a specific constructor.
A constructor that takes a reference (preferably const but not required) of the same class as an argument is just named the copy constructor because this is what it effectively does.

Answer 3

Besides the fact that it would not be a copy constructor and the compiler will generate the copy constructor unless you explicitly disable it, there is nothing to gain and much to loose. What is the correct semantics for a constructor out of a null pointer? What does this add to the user of your class? (Hint: nothing, if she wants to construct out of a heap object she can just dereference the pointer and use the regular copy constructor).