c++ - What happens to an STL iterator after erasing it in VS, UNIX/Linux?

Question

Please consider the following scenario:

map(T,S*) & GetMap(); //Forward decleration

map(T, S*) T2pS = GetMap();

for(map(T, S*)::iterator it = T2pS.begin(); it != T2pS.end(); ++it)
{
    if(it->second != NULL)
    {
        delete it->second;
        it->second = NULL;
    }
    T2pS.erase(it);
    //In VS2005, after the erase, we will crash on the ++it of the for loop.
    //In UNIX, Linux, this doesn't crash.
}//for

It seems to me that in VS2005, after the "erase", the iterator will be equal to end(), hence the crash while trying to increment it. Are there really differences between compilers in the behavior presented here? If so, what will the iterator after the "erase" equal to in UNIX/Linux?

Thanks...

Answer 1

Yes, if you erase an iterator, that iterator gets a so-called singular value, which means it doesn't belong to any container anymore. You can't increment, decrement or read it out/write to it anymore. The correct way to do that loop is:

for(map<T, S*>::iterator it = T2pS.begin(); it != T2pS.end(); T2pS.erase(it++)) {
    // wilhelmtell in the comments is right: no need to check for NULL. 
    // delete of a NULL pointer is a no-op.
    if(it->second != NULL) {
        delete it->second;
        it->second = NULL;
    }
}

For containers that could invalidate other iterators when you erase one iterator, erase returns the next valid iterator. Then you do it with

it = T2pS.erase(it)

That's how it works for std::vector and std::deque, but not for std::map or std::set.