Can inner classes access private variables? - c++

Question

class Outer {
class Inner {
public:
Inner() {}
void func() ;
};
private:
static const char* const MYCONST;
int var;
};
void Outer::Inner::func() {
var = 1;
}
const char* const Outer::MYCONST = "myconst";
This errors out when I compile with class Outer::Inner' has no member named `var'

Answer 1

An inner class is a friend of the class it is defined within.
So, yes; an object of type Outer::Inner can access the member variable var of an object of type Outer.
Unlike Java though, there is no correlation between an object of type Outer::Inner and an object of the parent class. You have to make the parent child relationship manually.
#include <string>
#include <iostream>
class Outer
{
class Inner
{
public:
Inner(Outer& x): parent(x) {}
void func()
{
std::string a = "myconst1";
std::cout << parent.var << std::endl;
if (a == MYCONST)
{ std::cout << "string same" << std::endl;
}
else
{ std::cout << "string not same" << std::endl;
}
}
private:
Outer& parent;
};
public:
Outer()
:i(*this)
,var(4)
{}
Outer(Outer& other)
:i(other)
,var(22)
{}
void func()
{
i.func();
}
private:
static const char* const MYCONST;
Inner i;
int var;
};
const char* const Outer::MYCONST = "myconst";
int main()
{
Outer o1;
Outer o2(o1);
o1.func();
o2.func();
}

Answer 2

An inner class has access to all members of the outer class, but it does not have an implicit reference to a parent class instance (unlike some weirdness with Java). So if you pass a reference to the outer class to the inner class, it can reference anything in the outer class instance.

Answer 3

Anything that is part of Outer should have access to all of Outer's members, public or private.
Edit: your compiler is correct, var is not a member of Inner. But if you have a reference or pointer to an instance of Outer, it could access that.

Answer 4

First of all, you are trying to access non-static member var outside the class which is not allowed in C++.
Mark's answer is correct.
Anything that is part of Outer should have access to all of Outer's members, public or private.
So you can do two things, either declare var as static or use a reference of an instance of the outer class to access 'var' (because a friend class or function also needs reference to access private data).
Static var
Change var to static If you don't want var to be associated with the instances of the class.
#include <iostream>
class Outer {
private:
static const char* const MYCONST;
static int var;
public:
class Inner {
public:
Inner() {
Outer::var = 1;
}
void func() ;
};
};
int Outer::var = 0;
void Outer::Inner::func() {
std::cout << "var: "<< Outer::var;
}
int main() {
Outer outer;
Outer::Inner inner;
inner.func();
}
Output- var: 1
Non-static var
An object's reference is a must to access any non-static member variables.
#include <iostream>
class Outer {
private:
static const char* const MYCONST;
int var;
public:
class Inner {
public:
Inner(Outer &outer) {
outer.var = 1;
}
void func(const Outer &outer) ;
};
};
void Outer::Inner::func(const Outer &outer) {
std::cout << "var: "<< outer.var;
}
int main() {
Outer outer;
Outer::Inner inner(outer);
inner.func(outer);
}
Output- var: 1
Edit - External links are links to my Blog.

Answer 5

var is not a member of inner class.
To access var, a pointer or reference to an outer class instance should be used. e.g. pOuter->var will work if the inner class is a friend of outer, or, var is public, if one follows C++ standard strictly.
Some compilers treat inner classes as the friend of the outer, but some may not. See this document for IBM compiler:
"A nested class is declared within the scope of another class. The name of a nested class is local to its enclosing class. Unless you use explicit pointers, references, or object names, declarations in a nested class can only use visible constructs, including type names, static members, and enumerators from the enclosing class and global variables.
Member functions of a nested class follow regular access rules and have no special access privileges to members of their enclosing classes. Member functions of the enclosing class have no special access to members of a nested class."