http://gcc.gnu.org/bugzilla/show_bug.cgi?id=45917
Summary: Friend of friend is allowed the access to the private
type through the template
Product: gcc
Version: 4.5.0
Status: UNCONFIRMED
Severity: normal
Priority: P3
Component: c++
AssignedTo: unassig...@gcc.gnu.org
ReportedBy: y...@tsoft.com
testcase below has struct R as private in class F.
class F declares class Q a friend, allowing it to see it's private members.
But operator<< is friend in Q, not in F.
Why is it allowed for operator<< to instantiate list<R> but not R r? This is a
bug.
--- testcase ---
#include <list>
#include <ostream>
using namespace std;
class F {
private:
struct R {
}; // R
friend class Q;
class Q {
list<R> l;
friend ostream& operator<<(ostream &os, const Q &q) {
// R r; // this breaks!
for (list<R>::const_iterator it = q.l.begin(); it != q.l.end(); it++) {
// this doesn't break! Why?
}
return os;
}
}; // Q
}; // F
