Re: Erasing in a vector while iterating through it

I have some code where there's this vector of pointers to objects and
I need to delete and erase some of them, the problem is that to know
which I need to iterate through the vector and I'm trying to do this
as efficient as possible.

The code is something like this:

struct Thing {
  int value;
  Thing* ptr;
  Thing() : ptr(0) { }
};

// .....

std::vector<Thing*> vec;

So the vector contains pointers to Thing-objects, and for some of
those objects ptr might be set, in which case it will point to one of
the Things pointed to by the pointers in the vector. What I need to do
is to go through the vector and for each Thing where value has a
specific value I need to delete the Thing pointed to by ptr and erase
it from the vector. One way I could do that is to go through the
vector in one pass and collect all those pointers to some other
container and then make another pass and erase them but I'd prefer to
do it in one pass and was wondering if that is legal (something like
this):

std::sort(vec.begin(), vec.end()); // Sort one the address of the
Things

for (size_t i = 0; i < vec.size(); ++i) {
  if (vec[i]->value == SOME_VALUE) {
    std::vector<Thing*>::iterator it =
std::::lower_bound(cells.begin(), cells.end(), vec[i]->ptr);
    vec[i]->value = 0;
    vec[i]->ptr = 0;
    delete *it;
    --i; // Decrement if the Thing removed is before i in vec
    vec.erase(it);
  }
}

Since vector::erase will invalidate all references and iterators to
elements after the one removed

there are two scenarios that I can
think of. The first is that the element is located after the i'th
element in which case I'll have to do some extra work (since I did --
i), or the element is before the i'th element (it can't be the i'th
element) in which case the i'th element (after the --i) should be the
same. Is this true?