why can't I access this method via inheritance?

Question

I have two simple classes, and wish to access the public method stuff by passing an int value. why can't I do that with an instance of Bar? Shouldn't it inherit the public method stuff. The type hinting gives the int a parameter, but it doesn't compile.

class Foo
{
public:
    int a;
    void stuff(int a){ std::cout << a << std::endl; }
};

class Bar : public Foo
{
protected:
    void stuff() { std::cout << "hello world"; }
};



void main()
{
    Bar b
    b.stuff(3);
}

try `b.Foo::stuff(3);`. You have to resolve explicitly the scope when calling the base class member function since the protected member function of the derived class hides it because they have the same name. — 101010, Jul 31 '14 at 18:09
that did the trick, thanks. Does this mean that all inherited objects carry about a hidden member of their base class and members/methods of which have to be accessed via #a cross between instance and static method calls? — user3791372, Jul 31 '14 at 18:13
I wouldn't say that they carry hidden members of their base class but rather an object of their base class. — 101010, Jul 31 '14 at 18:20

quantdev · Accepted Answer · 2014-07-31T18:22:07.687

6

Because Bar::stuff hides Foo::stuff (only the name matters when overload resolution is performed, parameters are ignored).

You can :

Bring it into sope with a using declaration
Or alternatively, explicitly qualify the call e.g. b.Foo::stuff(3);.

Note :

main() must return an int.

#include <iostream>

class Foo
{
public:
    int a;
    void stuff(int a){ std::cout << a << std::endl; }
};

class Bar : public Foo
{
public:
    using Foo::stuff;
protected:
    void stuff() { std::cout << "hello world"; }
};

int main()
{
    Bar b;
    b.stuff(3);
}

Or :

int main()
{
    Bar b;
    b.Foo::stuff(3);
}

edited Jul 31 '14 at 18:22

answered Jul 31 '14 at 18:16

quantdev

22,595
5
47
84

Thanks, this wraps everything up nicely. why must `main` return an int? Is it bad form in c++11 to declare main as void? – user3791372 Jul 31 '14 at 18:25
1

It is not related to c++11, see http://stackoverflow.com/questions/204476/what-should-main-return-in-c-and-c – quantdev Jul 31 '14 at 18:28
good to know - thanks – user3791372 Jul 31 '14 at 18:56

Alper · Answer 2 · 2014-07-31T19:13:41.610

In C++, name hiding can take place when one function in base class has the same name as one function in derived class.

Phases of the function call process

Name lookup
Overload resolution
Access control

Name lookup stops looking for other names as soon as it finds a name in derived class Bar. Therefore, Bar::stuff() hides any function with name stuff in Foo.

After the name lookup process,

overload resolution fails since there is no stuff(int) in Bar.
even though stuff() without int parameter is called, access control fails since stuff() is protected.

score 4 · Answer 3 · edited Jun 20 '20 at 09:12

4

When looking up the name stuff, the one in Bar found first. Once that is found, the access privileges are checked. Since Bar::stuff is protected, you are not able to use it from main.

From the draft standard:

10.2 Member name lookup [class.member.lookup]

1 Member name lookup determines the meaning of a name (id-expression) in a class scope (3.3.7). Name lookup can result in an ambiguity, in which case the program is ill-formed. For an id-expression, name lookup begins in the class scope of this; for a qualified-id, name lookup begins in the scope of the nestedname-specifier. Name lookup takes place before access control (3.4, Clause 11).

edited Jun 20 '20 at 09:12

Community

1
1

answered Jul 31 '14 at 18:10

R Sahu

196,807
13
136
247

I thought the whole point of using access modifiers like public/protected/private was to set the visibility so that the `stuff` in Bar isn't found? – user3791372 Jul 31 '14 at 18:11
@user3791372, please see my updated answer. – R Sahu Jul 31 '14 at 18:16
@user3791372: The point is to set the visibility so that the `stuff` in `Bar` can't be _used_. – Mooing Duck Jul 31 '14 at 18:17

score 0 · Answer 4 · answered Jul 31 '14 at 18:16

0

b.Foo::stuff(3);

Or, you can do a trick to access any protected:)

void main()
{
    Bar b;
    class BarAccess : public Bar { friend void main(); }
    BarAccess * ba = reinterpret_cast<BarAccess *>(&b);
    ba->stuff(3);
    ba->stuff();
}

answered Jul 31 '14 at 18:16

yshurik

763
5
12

that's some 'trick' ;) – user3791372 Jul 31 '14 at 18:22
2

No offence, but I don't agree with this trick. You discard the meaning of encapsulation doing this. – 101010 Jul 31 '14 at 18:25

why can't I access this method via inheritance?

4 Answers4