I wanted to pass-in functions as arguments to a templated function without having any indirection. To achieve this I created two nested structs, each defining the function I wish to pass in to the templated function. Each of the structs accesses data members from the outside class B:
namespace A{
class B{
public:
B();
template <typename T>
void templatedFunction(T t){
//I pass one of the struct objects in to here, to invoke the desired function
t();
}
private:
struct X{
void operator(){
do();
}
void do(){
//Accesses the data members of class B
e->doSomething();
}
};
struct Y{
void operator(){
do();
}
void do(){
//Accesses the data members of class B
d.doSomething();
}
};
C* c;
D d;
E* e;
};
}
and the compiler errors I get are pretty much all of the format:
error: invalid use of non-static data member B::d
for the lines within the struct accessing the class data members and on the lines declaring the data members in B.
A nested class in C++ does not (automatically) have access to an instance of the containing class. It's just a class definition like any other. You need a B instance to access non-static data members of B.
You can restructure your nested class Y as
struct Y
{
void operator()( B& b ){
do( b );
}
void do( B& b ){
//Accesses the data members of class B
b.d.doSomething();
}
};
and fix your function template and calls, and class X, accordingly.
Note that your presented code for operator(), with no argument list, would not have compiled, but I'm not downvoting since you are stopped by another compilation error (i.e., possibly it is the real code you're showing).
If you have a struct (or class for that matter) nested in another class, it it not treated specially in any way. It works exactly the same as if the struct was defined outside of the enclosing class. The only thing which is different is the scope of nested class name. So, if you have
class A { class B{}; };
and
class B{}; class A {};
the only difference is that outside class A you need to name the class B as A::B in the first case, and just B in the second class. There is no other difference, and in particular class B does not gain any special access to class A members.
Related
Why can I not access the protected members from a template method of a class?
I may be missing some special friend declaration here but it eludes me. I feel like I should be able to do this.
The error is:
error: ‘char* ClassB<char>::a’ is protected
Example source:
template<typename T>
class ClassA;
template<typename T>
class ClassB {
protected:
T* a;
public:
ClassB()
: a(0) {}
template<typename U>
ClassB(const ClassB<U>& other)
:
// error: ‘char* ClassB<char>::a’ is protected
a(other.a) {}
};
////
template<typename T>
class ClassA : public ClassB<T> {
public:
ClassA() {}
};
////
namespace name {
typedef ClassA<char> A;
typedef ClassB<const char> B;
}
int main() {
name::A a;
name::B b = a;
return 0;
}
You can't do it for the same reason that ClassA cannot access the protected/private members of ClassB. The fact that templated classes share a common name doesn't really matter to them. ClassB<T> and ClassB<U> treat each other like entirely different classes and so their members aren't accessible to each other.
The reason for this becomes clearer when you realize you can specialize templated classes, which means that it's possible to have implementations of ClassB that do not have a member named a (or do have a member named a, but use it in an entirely different way, and so it shouldn't be accessed).
The fact is that ClassB<T> and ClassB<U> are different classes (unless T = U, but that's not generally true and so the compiler can't rely on that assumption). Hence, they can't access each other.
This is an abstraction feature of the language. Just like two independent classes, ClassB<T> and ClassB<U> treat each other like they are not related and are different classes and so they cannot access private and protected members of each other.
I have class A which has to implement some functions. Since implementing one of them needs it's own data structures, I assumed A contain another class B, which has all needed data structures and functions. However, B is also need to use data structures and functions of A, as well. I used two classes calling each others using forward declaration. But there is still problems. For example, I need to make all data structures in A public, in order to B can access it. I tried using friend classes, but when I declare B as an abstract classes with sub-classes which implements B's functionalities, I need to make all data structures of A, as public. Because friend class doesn't work for inherited sub-classes, all data structures of A, needs to be public. This makes my design quite messy.
class B;
class A{
protected:
int ds[100];
B * b;
public:
a_func(){
b->b_func();
}
};
class A;
class B{
A * a;
public:
b_func(){
a->a_func();
}
};
class sub_B:public B{
public:
b_func(){
a->a_func();
a->ds ...;
}
}
My question is: is there any alternative design?
I also tried making A an abstract class and class B implements a function of it, however, it doesn't conceptually makes sense to build an object of B, when I want an object of A.
You don't have to provide member function definitions inside a class definition:
class A;
class B;
class A {
// no need for public
B * b;
void a_funct(void);
};
class B {
// no need for public here, too
A * a;
void b_funct(void);
};
// the following goes in a source file,
// otherwise you should mark it as inline
void A::a_funct() {
b->b_funct();
}
void B::b_funct() {
a->a_funct();
}
Note that above code serves only as example, in its current shape it's nothing but a fancy endless (recursion) loop.
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 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.
Is there anyway to declare an object of a class before the class is created in C++? I ask because I am trying to use two classes, the first needs to have an instance of the second class within it, but the second class also contains an instance of the first class. I realize that you may think I might get into an infinite loop, but I actually need to create and instance of the second class before the first class.
You can't do something like this:
class A {
B b;
};
class B {
A a;
};
The most obvious problem is the compiler doesn't know how to large it needs to make class A, because the size of B depends on the size of A!
You can, however, do this:
class B; // this is a "forward declaration"
class A {
B *b;
};
class B {
A a;
};
Declaring class B as a forward declaration allows you to use pointers (and references) to that class without yet having the whole class definition.
You can't declare an instance of an undefined class but you can declare a pointer to one:
class A; // Declare that we have a class A without defining it yet.
class B
{
public:
A *itemA;
};
class A
{
public:
B *itemB;
};
There's an elegant solution using templates.
template< int T > class BaseTemplate {};
typedef BaseTemplate< 0 > A;
typedef BaseTemplate< 1 > B;
// A
template<> class BaseTemplate< 0 >
{
public:
BaseTemplate() {} // A constructor
B getB();
}
// B
template<> class BaseTemplate< 1 >
{
public:
BaseTemplate() {} // B constructor
A getA();
}
inline B A::getB() { return A(); }
inline A B::getA() { return B(); }
This code will work! So, why does it
work? The reason has to do with how
templates are compiled. Templates
delay the creation of function
signatures until you actually use the
template somewhere. This means that
neither getA() nor getB() will have
their signatures analyzed until after
both classes A and B have already been
fully declared. That's the magic of
this method.
Is this close to what you want: The first class contains the second class, but the second class (that is to be created first) just has a reference to the first class?
This is called cross reference. See here an example.