which type should "std::set::begi n() const" return?

[snnn]

On the book <Generic Programming and the STL>( Matthew . H . Austern ),this function is defined as
iterator set::begin() const.
However, why should a const object returns a non-const iterator?
Then, I found, in this book, the semantic of set::iterator is defined as same as set::const_iter ator. Both of them must be const!
I tried to read the source of GNU STL(version 3.4.1).They were using a red-black tree to implant it (std::set has a _RBtree.) .Both set::iterator and set::const_iter ator are simply defined as _RBtree::const_ iterator.
But, in VC++, it is deferent.
There is a base class named xtree which also a red-black tree. Then the class map and set inherit from xtree (which implantation is better? is-a? has-a?).They don't defined iterator and const_iterator theirself,just inherit them from the base class xtree,and the member function begin() is defined as :
const_iterator set::begin() const;
iterator set::begin().
So ,we can initialize a mutable set, get the begin of it, then try to modify it!
The following code has been compiled successful under VC++2003
int a[4]={1,2,3,4};
std::set<int> s(a,a+4);
std::set<int>:: iterator i=s.begin();
(*i) = 9;
std::cout<<"Now s = ";
std::copy(s.beg in(),s.end(),st d::ostream_iter ator<int>(std:: cout,","));
std::cout<<std: :endl;
std::set<int>:: const_iterator p=s.find(9);
if(p != s.end() )
std::cout<<(*p) ;
else std::cout<<"Can not find the element special"<<std:: endl;

A set must be a sorted container, but now it is not! So anything terrible things it would happen.
Is this a bug of VC++?

A bug of VC++.net STL
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[snnn]

!!!
You are right!

[ivy]

Have you got the rigth answer??
I've been thinking about it since we last meet.