Двоичный поиск с использованием итераторов, почему мы используем "(end-begin)/2"?

Я изучаю итераторы и задерживался на 3 дня, выясняя, почему мы используем:

auto mid = text.begin() + (end - beg) / 2;

код:

int main()

{
    vector<int> text{ 10,9,8,7,6,5,4,3,2,1 };
    int sought = 3;
    // text must be sorted
    // beg and end will denote the range we're searching
    auto beg = text.begin(), end = text.end();
    auto mid = text.begin() + (end - beg) / 2; // original midpoint
                                               // while there are still elements to look at and we haven't yet found sought
    while (mid != end && *mid != sought) {
        if (sought < *mid) // is the element we want in the first half?
            end = mid; // if so, adjust the range to ignore the second half
        else // the element we want is in the second half
            beg = mid + 1; // start looking with the element just after mid
        mid = beg + (end - beg) / 2;// new midpoint
    }

    system("pause");
}

почему

auto mid = text.begin() + (end - beg) / 2;

и не:

auto mid = text.begin() + text.size() / 2;

Пожалуйста, помогите.

Ответы

So what the best way to fix the bug? Here one way:
 6:             int mid = low + ((high - low) / 2);

Probably faster, and arguably as clear is:
 6:             int mid = (low + high) >>> 1;

In C and C++ (where you don't have the >>> operator), you can do this:
 6:             mid = ((unsigned int)low + (unsigned int)high)) >> 1;