Confused about copy constructor...

Hi,
I need help about copy constructor

Code:

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class A {
public:
    ...
    A(const A& rhs) {_n1 = rhs._n1; _n2 = rhs._n2;}
private:
    int _n1;
    int _n2;
}

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Compiler just gave warning about using private data member of rhs;
I know that it doesn't need to explicit declare copy constructor in this case.
But what if I want to declare it explicitly, how should I do? Removing the private access label seems not a good idea for me.
Thx for reply

What is the warning?

here's the code

Code:

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22    Interval(const Interval& rhs) {
 23      _min = rhs._min;
 24      _max = rhs._max;
 25    }

 26    Interval& operator= (const Interval& rhs) {
 27      _min = rhs._min;
 28      _max = rhs._max;
 29      return (*this);
 30    }

---

Code:

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  2 error: ‘int Interval::_max’ is private
  3 Interval.cpp:25: error: within this context
  4 Interval.h:77: error: ‘int Interval::_min’ is private
  5 Interval.cpp:25: error: within this context

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Can you post a compilable example that demonstrates the problem?

Well, forget aout above codes.
I try to use small compilable case to test, and it works.
Now my question is that.
If it works, why?
in copy constructor the arguments const Type& rhs
why can we directly access member of rhs without using public methods?

Since it works, it's obvious that what you are doing is "correct". The error you are getting stems somehow from something in your code and not from the copy constructor code.
We can implicitly access private members of an external instance of the same type. The why is probably to be able to write copy constructors.
Now, if you're doing templated code, you should be aware that if you have a class of type A<T> and try to access members of A<S> inside A<T>, you will get a compile error.
The solution would be to make A<T> a friend of A<S>.

Quote:

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Originally Posted by ovid

Well, forget aout above codes.
I try to use small compilable case to test, and it works.
Now my question is that.
If it works, why?
in copy constructor the arguments const Type& rhs
why can we directly access member of rhs without using public methods?

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If the code you have supplied is representative of what you are doing, then it works because it is supposed to.

Access control (private, protected, public) works on a per-class basis: a member function of a class can access the private members of any instance of that class.

The real question is: what haven't you shown that you did differently, that caused the compilation error?

My guess - despite the fact you haven't actually informed us - is that your actual example was a template class.

If X and Y are different classes, and SomeClass is a template class, the member functions of SomeClass<X> are unable to access private members of SomeClass<Y> unless SomeClass<Y> specifically declares SomeClass<X> as a friend.

Quote:

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Originally Posted by Elysia

Can you post a compilable example that demonstrates the problem?

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Bad question; his point is that it doesn't compile!

In any case ovid, how could we possibly know what mistake exists in the real code without it being posted here?
You either want an answer and you post the relevant real code, or you don't want an answer and you just jerk people around playing guessing games forever. Your choice.

Quote:

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Originally Posted by iMalc

Bad question; his point is that it doesn't compile!

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The originally supplied code compiles just fine, as it should.
So I asked for some real code that demonstrates the problem.