Re: Why friendship not applicable to function arguments

From:
"Kaz Kylheku" <kkylheku@gmail.com>
Newsgroups:
comp.lang.c++
Date:
29 Jun 2006 14:11:14 -0700
Message-ID:
<1151615474.868932.177930@m73g2000cwd.googlegroups.com>
Ninan wrote:

When compiling using gcc 2.95.3, I get the following error for the
code snippet posted below.
bash-2.03$ g++ membAccess.C
membAccess.C: In function `int main()':
membAccess.C:13: `struct A B::m_a' is private
membAccess.C:38: within this context
Is this how it is supposed to work or is this a compiler specific issue

   1 #include <iostream>
      2
      3
      4 struct A
      5 {
      6 int y;
      7 };
      8
      9 void useAalter (A* pA);
     10 class B
     11 {
     12
     13 A m_a;
     14 public:
     15 B (A& a)
     16 {
     17 m_a = a;
     18 }
     19 A* getA () { return &m_a;}
     20 friend void useAalter(A*pA);
     21 };


Because useAalter is a friend of class B, it is allowed to access
private members in that class.

     28 void useAalter (A* pA)
     29 {
     30 std::cout << pA->y << std::endl;
     31 }


But useAalter does not actually access anything in class B. This
function makes no use of its frienship to B whatsoever. It accesses the
identifier A::y. This is allowed because y is public in struct A.

     32
     33 int main ()
     34 {
     35 A a = {45};
     36 B bInst (a);
     37 useA (bInst.getA());
     38 useAalter (&(bInst.m_a)) ;
     39 }


Here, it is the function main which accesse the name B::m_a in line 38.
Since main isn't a friend of the class B and m_a is private, this
doesn't work.

Access protection is concerned with name lookups, not pointer
dereferences.

The function useAalter has no idea that the A * which it is working
with is in fact B::m_a. It could be any A struct whatsoever:

   A a = { 45 };

   useAalter(&a); // nothing to do with any B object

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