Reputation: 21
Suppose I have the following class:
class Test
{
int num;
public:
Test(int x):num(x){}
Test(const Test &rhs):num(rhs.num+1){}
};
int main()
{
Test test(10);
Test copy = test;
}
The num
in the copy should be 11
, and my question is about inside the copy constructor, why can we access the private member num
of test
using num
to initialize the num
in the copy? What is confusing me is that if you type cout<<test.num<<endl
, of course it's wrong because you are trying to access the private num
, but if you pass the test by reference to the copy constructor, it works, can anybody tell me what is going on here?
Upvotes: 1
Views: 3047
Reputation: 40613
Access limitations are per-class, not per-object.
"private" means -- can only be accessed from within the same class.
"protected" means -- can be accessed from within the same class, and can also be accessed from within derived-classes (in derived classes, protected non-static members can only be accessed through variables with derived class type).
"public" means -- can be accessed by anything.
The point of access limitations is to limit the region of code that has to be inspected in order to understand where the values are used, rather than to stop code from using the values.
Upvotes: 3
Reputation: 89946
private
doesn't mean private to the object instance. It means private to that class. An instance of a class T
can access private members of other instances T
. Similarly, a static method in a class T
can access private members of instances of T
.
If private
restricted access to only the individual instance, it would make objects non-copyable, since as you pointed out, the copy constructor would not be able to read data from the original instance.
Upvotes: 1
Reputation: 571
Private members are private to the class itself, not the instances of the class.
Upvotes: 6