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.x is A's private member, how come B can refer to it (even it's a const) ?
Look up a C++ Reference and learn How To Ask Questions The Smart WayOriginally Posted by Bjarne Stroustrup (2000-10-14)
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?
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.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.
Look up a C++ Reference and learn How To Ask Questions The Smart Way
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?
