x is A's private member, how come B can refer to it (even it's a const) ?Code:class B{
public:
B(){}
void check(const int& y){cout<<y<<endl;}
};
class A{
int x;
public:
A(){x=1;}
void foo(){B ob; ob.check(x);}
};
int main(){
A oa;
oa.foo();
return 0;
}
B doesn't refer to A's private variable. All it sees is a parameter passed into a function. A can see its own private variable, and can choose to send it wherever it wants.
You are passing an argument, so the B object never actually deals with the member variable named x of A, but its check member function deals with a local variable named y. It so happens that pass by reference is used, so y becomes an alias for x when ob.check(x) is called.Quote:
x is A's private member, how come B can refer to it (even it's a const) ?
OK, To make my question more clear, I modify the code:
Now you see before( and after) the call, A's private member variable actually was changed.Code:class B{
public:
B(){}
void check( int& y){ y++;}
};
class A{
int x;
public:
A(){x=1;}
void foo(){
cout<<x<<endl;
B ob;
ob.check(x);
cout<<x<<endl;}
};
int main(){
A oa;
oa.foo();
return 0;
}
Well, apparently class A wanted it changed, since it sent the variable off to a function pass-by-reference. The responsibility for changing the value rests right there, not in class B.
Thanks. But x is A's private variable, how would it be changed by a non-friend function?
Quote:
Originally Posted by tabstop
If you pass a value by reference it does NOT matter who owns it, what class it is in. If you pass a variable by reference it is able to be changed by the function regardless of the ownership.
"Private" means "Nobody can see this name," not "Nobody can touch this variable."
I think you have a fundamental misunderstanding of scope and access control.Quote:
But x is A's private variable, how would it be changed by a non-friend function?
Let's ask a few questions and see how you answer them:
1. Is there a local variable named y in foo(int&)?Code:void foo(int& x)
{
x = 0;
}
void bar()
{
int y = 1;
foo(y);
}
class Z
{
private:
void baz()
{
int z = 2;
foo(z);
}
};
2. Is there a local variable named z in foo(int&)?
3. Remove the call of foo(y) in bar(). Now, without making any further modification to bar(), can you modify foo(int&) such that it changes the value of y in bar()?
4. Remove the call of foo(z) in Z::baz(). Now, without making any further modification to Z::baz(), can you modify foo(int&) such that it changes the value of z in Z::baz()?
The answer to all four questions is no. So, foo(int&) does not have access to y or z. In particular, foo(int&) does not have access to z, but due to pass by reference its parameter x can be an alias (another name) for z if Z::baz() calls foo(z). So, the private access is respected since only a member function (or friend) of Z can access z.
Your ATM card and pin are private to you.
If you choose to give me your card and pin #, what's to stop me from emptying your bank account?