why isnt a derived class not permissioned to access its protected base class member when its templatized?
class MyBase {
protected:
int foo;
};
template<typename Impl>
class Derived : public Impl {
public:
int getfoo() {
return static_cast<Impl*>(this)->foo;
}
};
The compiler complains foo is protected. why?
error: int MyBase::foo is protected
You are accessing foo through an MyBase* instead of a Derived<MyBase>*. You may only access protected members through your own type, not through the base type.
Try this instead:
int getfoo() {
return this->foo;
}
From the C++ 2003 standard, 11.5/1 [class.protected]: "When a friend or a member function of a derived class references a protected nonstatic member function or
protected nonstatic data member of a base class … the access must be through a
pointer to, reference to, or object of the derived class itself (or any class derived from that class)"
Related
This question already has an answer here:
Why can I call base template class method from derived class
(1 answer)
Closed 6 years ago.
template<int N>
struct B
{
protected:
void f() {}
};
template<int N>
struct A : B<N>
{
A()
{
this->f(); // ok
f(); // error : use of undeclared identifier 'f'
}
};
int main()
{
A<8> a;
}
My C++ compiler is clang 3.8.
Why doesn't clang allow a derived class to call a protected base function?
This is nothing specific to clang but C++ access control and protected access specifier.
The language specification wants to ensure that you are accessing a protected member of some base subobject that belongs to the derived class object. You are not supposed to be able access protected members of some unrelated independent objects of base type.
For this reason, you have to access protected members through pointer->member syntax, reference.member or object.member syntax, where the pointer/reference/object refers to the derived class.
This is exactly inline with what you have shown in your example:
template<int N>
struct B
{
protected:
void f() {}
};
template<int N>
struct A : B<N>
{
A()
{
this->f(); // ok
f(); // error : use of undeclared identifier 'f'
}
};
Link:
Protected members are not as private as private members, which are accessible only to members of the class in which they are declared, but they are not as public as public members, which are accessible in any function.
Protected members that are not declared as static are accessible to friends and member functions in a derived class only through a pointer to, reference to, or object of the derived class.
I have the following class structure:
class Base {
public:
std::set<Index> openIndices;
Base() {};
};
template<typename lhs_t, typename rhs_t>
class Derived : public Base {
public:
lhs_t &lhs;
rhs_t &rhs;
Derived(lhs_t &_lhs, rhs_t &_rhs) :
Base(),
lhs(_lhs),
rhs(_rhs),
openIndices(std::set_symmetric_difference(lhs.openIndices, rhs.openIndices))
{
}
};
So basicly a template Class Derivedderived from the base class Base. When accessing the member variables ob the baser class I get the following error:
test.h:34:88: error: class ‘Derived<lhs_t, rhs_t>’ does not have any field named ‘openIndices’
I am aware of this question: I cannot access the member variables in case my class is derived from a template class. But in my case I am not derived from a template class, still I can't access the member variables. Could anyone tell me why?
You cant initialize member variables of a base class. You have to provide an appropriate constructor in the base class and call this constructor.
It seems that a protected member from a template policy class is inaccessible, even with a class hierarchy which seems correct.
For instance, with the following code snippet :
#include <iostream>
using namespace std;
template <class T>
class A {
protected:
T value;
T getValue() { return value; }
public:
A(T value) { this->value = value; }
};
template <class T, template <class U> class A>
class B : protected A<T> {
public:
B() : A<T>(0) { /* Fake value */ }
void print(A<T>& input) {
cout << input.getValue() << endl;
}
};
int main(int argc, char *argv[]) {
B<int, A> b;
A<int> a(42);
b.print(a);
}
The compiler (clang on OS X, but gcc returns the same type of error) returns the following error :
Untitled.cpp:18:21: error: 'getValue' is a protected member of 'A<int>'
cout << input.getValue() << endl;
^
Untitled.cpp:25:5: note: in instantiation of member function 'B<int, A>::print' requested here
b.print(a);
^
Untitled.cpp:8:7: note: can only access this member on an object of type 'B<int, A>'
T getValue() { return value; }
^
1 error generated.
The strange thing is that the last note from the compiler is totally correct but already applied since the b object is of type 'B<int, A>'. Is that a compiler bug or is there a problem in the code ?
Thanks
You have misunderstood the meaning of protected access.
Protected members are callable by derived classes. But only on the base object contained inside the class itself.
For example, if i simplify the problem, using :
class A {
protected:
void getValue(){}
};
class B : protected A
{
public:
void print(A& input)
{
input.getValue(); //Invallid
}
};
getValue cannot be called on a "A" object other than the "A" object inside the class itself.
This for example is valid.
void print()
{
getValue(); //Valid, calling the base class getValue()
}
As pointed out by Dan Nissenbaum and shakurov. This is however also valid:
void print(B& input)
{
input.getValue();
}
This is because we explicitly say that input is a object of B. And the compiler know that all that objects of B has protected access to getValue. In the case when we pass a A&, the object might as-well be a type of C, wich could be derrived from A with private access.
Let's forget for a minute about the template and look at this:
class A {
protected:
int value;
int getValue() { return value; }
public:
A(int value) { this->value = value; }
};
class B : protected A {
public:
B() : A(0) { /* Fake value */ }
void print(A& input) {
cout << input.getValue() << endl;
}
};
The print() method's implementation is wrong because you can't access non-public member of A inside B. And here's why: from within B, you can only access non-public members of B. Those members may be either inherited or not — it doesn't matter.
On the other hand, A& input may not be a reference to an instance of B. It may be a reference to another subclass (which may well have getValue() inaccessible).
Member functions of a derived class have access to protected base class members within any object of its type that is passed as an argument so long as the explicitly declared class of the object passed as an argument is that of the the derived class (or a further derived class).
Objects explicitly passed as the base class type cannot have their protected members accessed within the derived class's member functions.
In other words, if we have:
class A
{
protected:
int x;
}
class B : public A
{
void foo(B b)
{
b.x; // allowed because 'b' is explicitly declared as an object of class B
}
void goo(A a)
{
a.x; // error because 'a' is explicitly declared as having *base* class type
}
};
...then the line a.x is not allowed because the explicit type of the argument is A, but the rule for protected access only applies to objects explicitly defined as the same class as the class attempting to access the member. (...Or a class derived from it; i.e., if class Cderives from B, then passing an object explicitly declared as an object of class C will also have x accessible within B member functions.)
The reason for this is given by shakurov when he writes (paraphrasing)
A& input might not be a reference to an instance of B. It may be a
reference to another subclass (which may well have getValue()
inaccessible)
An excellent explication of this answer is also given here: accessing a protected member of a base class in another subclass.
As a matter of interest, I believe that this comes from the C++ standard here:
11.4 Protected member access [class.protected] 1 An additional access check beyond those described earlier in Clause 11 is applied when a
non-static data member or non-static member function is a protected
member of its naming class (11.2)115 As described earlier, access to a
protected member is granted because the reference occurs in a friend
or member of some class C. If the access is to form a pointer to
member (5.3.1), the nested-name-specifier shall denote C or a class
derived from C. All other accesses involve a (possibly implicit)
object expression (5.2.5). In this case, the class of the object
expression shall be C or a class derived from C.
Don't get distracted by the template. It has nothing to do with the error. The line in main that the compiler is complaining about creates an object of type B<int, a> and tries to access a protected member. That's not legal, regardless of the type. You can only use protected members from inside a member function or friend function. For example:
struct S {
protected:
void f();
};
int main() {
S s;
s.f(); // error: attempts to call a protected member function
}
I believe, a derived class can override only those functions which it inherited from the base class. Is my understanding correct.?
That is if base class has a public member function say, func, then the derived class can override the member function func.
But if the base class has a private member function say, foo, then the derived class cannot override the member function foo.
Am i right?
Edit
I have come up with a code sample after studying the answers given by SO members. I am mentioning the points which i studied as comments in the code. Hope i am right. Thanks
/* Points to ponder:
1. Irrespective of the access specifier, the member functions can be override in base class.
But we cannot directly access the overriden function. It has to be invoked using a public
member function of base class.
2. A base class pointer holding the derived class obj's address can access only those members which
the derived class inherited from the base class. */
#include <iostream>
using namespace std;
class base
{
private:
virtual void do_op()
{
cout << "This is do_op() in base which is pvt\n";
}
public:
void op()
{
do_op();
}
};
class derived: public base
{
public:
void do_op()
{
cout << "This is do_op() in derived class\n";
}
};
int main()
{
base *bptr;
derived d;
bptr = &d;
bptr->op(); /* Invoking the overriden do_op() of derived class through the public
function op() of base class */
//bptr->do_op(); /* Error. bptr trying to access a member function which derived class
did not inherit from base class */
return 0;
}
You can override functions regardless of access specifiers. That's also the heart of the non-virtual interface idiom. The only requirement is of course that they are virtual.
But if the base class has a private member function say, foo, then the derived class cannot override the member function foo.
In Java, you can't. In C++, you can.
You are correct in that to override a function in a derived class, it must inherit it from the base class (in addition, the base class function must be virtual). However you are wrong about your assumption that virtual functions are not inherited. For example, the following works well (and is actually a known idiom for precondition/postcondition checking):
class Base
{
public:
void operate_on(some thing);
private:
virtual void do_operate_on(some thing) = 0;
};
void Base::operate_on(some thing)
{
// check preconditions
do_operate_on(thing);
// check postconditions
}
class Derived: public Base
{
// this overrides Base::do_operate_on
void do_operate_on(some thing);
};
void Derived::do_operate_on(some thing)
{
// do something
}
int main()
{
some thing;
Base* p = new Derived;
// this calls Base::operate_on, which in turn calls the overridden
// Derived::do_operate_on, not Base::do_operate_on (which doesn't have an
// implementation anyway)
p->operate_on(thing);
delete p;
}
A way to see that private methods are really inherited is to look at the error messages generated by the following code:
class Base
{
private:
void private_method_of_B();
};
class Derived:
public Base
{
};
int main()
{
Derived d;
d.private_method_of_B();
d.method_that_does_not_exist();
}
Trying to compile this with g++ leads tot he following error messages:
privatemethodinheritance.cc: In function 'int main()':
privatemethodinheritance.cc:4: error: 'void Base::private_method_of_B()' is private
privatemethodinheritance.cc:15: error: within this context
privatemethodinheritance.cc:16: error: 'class Derived' has no member named 'method_that_does_not_exist'
If class Derived wouldn't inherit that function, the error message would be the same in both cases.
No You can not over ride any function in base class .
My reason for this notion is that if you define the function in derived class that has the same function signiture in base class , the base class function will become hidden for derived class .
For more information about this exciting issue just visit :
http://publib.boulder.ibm.com/infocenter/comphelp/v8v101/index.jsp?topic=%2Fcom.ibm.xlcpp8a.doc%2Flanguage%2Fref%2Foverload_member_fn_base_derived.htm
It's also depends on the type of inheritance
class Derived : [public | protected | private] Base { };
In order to override a function inherited from base class, it should be both virtual, and classified as public or protected.
I have abstract class A, from which I inherit a number of classes. In the derived classes I am trying to access protected function in A trough A pointer. But I get a compiler error.
class A
{
protected:
virtual void f()=0;
};
class D : public A
{
public:
D(A* aa) :mAPtr(aa){}
void g();
protected:
virtual void f();
private:
A* mAPtr; // ptr shows to some derived class instance
};
void D::f(){ }
void D::g()
{
mAPtr->f();
}
The compiler error says : cannot access protected member A::f declared in class A.
If I declare mAPtr to be D*, instead A* everything compiles. And I don't understand why is this.
Relying on private access works on unrelated instances of the same type.
Relying on protected access works on unrelated instances of the same type (and of more derived types).
However, relying on protected access does not work on unrelated instances of a base type.
[n3290: 11.5/1]: When a friend or a member function of a derived
class references a protected nonstatic member function or protected
nonstatic data member of a base class, an access check applies in
addition to those described earlier in clause 11. Except when forming
a pointer to member (5.3.1), the access must be through a pointer
to, reference to, or object of the derived class itself (or any class
derived from that class) (5.2.5). If the access is to form a pointer
to member, the nested-name-specifier shall name the derived class (or
any class derived from that class).
So D or something derived from D, but not A.
It's an oft-questioned cute oddity about C++ that nonetheless is designed to try to avoid pitfalls. After all, you don't know what type *mAPtr really has.
A class containing a protected section means that this class allows derived classes to manipulate their base class in any way they choose (as far as the protected interface allows).
Class D objects can manipulate their own A part. In doing say they probably want to maintain some invariants.
Suppose there is (or will be in the future!) another class E, also inherited from A. Class E objects also can manipulate their own A part, and they may be enforcing different invariants.
Now, if a class D object was allowed to manipulate the A part of any object, it can't ensure the invariants. A D object may do something to the A part of an E object that breaks that E object. That's why it is not allowed.
But if you really want to, perhaps a way to call A::f, without exposing it to everybody, would be via a friend function.
class A;
namespace detail
{
void call_f(A*);
}
class A
{
friend void detail::call_f(A*);
private:
virtual void f() = 0;
};
namespace detail
{
void call_f(A* a) { a->f(); }
}
class D: public A
{
public:
void g() { detail::call_f(mAPtr); }
private:
void f() {}
A* mAPtr;
};
This relies on users being disciplined enough to stay out of namespaces whose name clearly indicates that it contains implementation details.
You forgot using ; after class declaration:
class A
{
protected:
virtual void f()=0;
};
class D : public A
{
public:
void g();
protected:
void f();
private:
A* mAPtr; // ptr shows to some derived class instance
};
Besides, you don't need to store base class pointer.