The termination conditions about the binary search

About some thoughts of binary search:
To avoid some confusion of the binary search, like dead-loop or something.
Every time i write binary search, i always use the 'while' condition like this:

int l = 0;
int r = nums.size()-1;

while (l < r-1) {
// insert the specific code
} 

And then to check both the nums[l] and nums[r] with the final result.
This way could definitely eliminate the possibility of  dead-loop, like 'l' always be the 'mid', and 'mid' always equals to 'l'.
But this way is not very elegant, so, when could we use the condition like:

while (l

To do like this way, we need have condition like:

while (l < r) {
            int mid = l + (r - l) / 2;
            if (citations[mid] >= size-mid) r = mid;
            else l = mid + 1; // The key
}

So if, l = mid + 1, then we could use while (l < r).
Like the following problem: https://leetcode.com/problems/h-index-ii/
Both solution can reach the correct answer:

// First Edition
class Solution {
public:
    int hIndex(vector& citations) {
        if ((int)citations.size() == 0) return 0;
        int size = (int)citations.size();
        int l = 0;
        int r = size - 1;
        while (l < r-1) {
            int mid = l + (r - l) / 2;
            if (citations[mid] >= size-mid) r = mid;
            else l = mid + 1;
        }
        if (citations[l] >= size - l) return size-l;
        else if (citations[r] >= size - r) return size-r;
        return 0;
    }
};

// We can also write like this
class Solution {
public:
    int hIndex(vector& citations) {
        if ((int)citations.size() == 0) return 0;
        int size = (int)citations.size();
        int l = 0;
        int r = size - 1;
        while (l < r) {
            int mid = l + (r - l) / 2;
            if (citations[mid] >= size-mid) r = mid;
            else l = mid + 1;
        }
        if (citations[r] != 0)
            return size-r;
        return 0;
    }
};