Why C++ copy constructor must use const object?

I understand that when we define a class copy constructor of the class is necessary as Rule of three states. I also notice that the argument of the copy constructor is usually const as the following codes illustrate:

class ABC {
public:

    int a;
    int b;
    ABC(const ABC &other)
    { 
        a = other.a;
        b = other.b;
    }
}

My question is what would happen if the argument of the copy constructor is not const:

class ABC
{
    public:
    int a;
    int b;
    ABC(ABC &other)
    { 
        a = other.a;

        b = other.b;
    }
}

I understand that in some cases if the argument of the copy constructor is const then the second implementation would fail. Also if the argument of the copy constructor is const then the object to be copied will not change its content during the process. However, I do notice that some people still use the second implementation rather than the first one. Are there any reasons that the second implementation is preferred?

Answer

• Logically, it should make no sense to modify an object of which you just want to make a copy, though sometimes it may have some sense, like a situation where you'd like to store the number of time this object has been copied. But this could work with a mutable member variable that stores this information, and can be modified even for a const object (and the second point will justify this approach)




• You would like to be able to create copy of const objects. But if you're not passing your argument with a const qualifier, then you can't create copies of const objects...





• You couldn't create copies from temporary reference, because temporary objects are rvalue, and can't be bound to reference to non-const. For a more detailed explanation, I suggest Herb Sutter's article on the matter