First occurrence in a binary search

An addition to Jon Skeets post:

The potential faster implementation is actually not hard to implement and adds only 2 lines of code, here is how I’d do it:

    if (midVal < key)
        low = mid + 1;
    else if (midVal > key)
        high = mid - 1;
    else if (low != mid) //Equal but range is not fully scanned
        high = mid; //Set upper bound to current number and rescan
    else //Equal and full range is scanned
        return mid;