Dichotomy queries values in an array

public class Main {
    public static void main(String[] args) {
        // An ordered array , There are no repeating elements
        int[] nums = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 78, 788, 878, 1000 };
        // Declare intermediate value
        int middle = 0;
        // Declare the start value
        int start = 0;
        // The value found is 3, Find the value in this array as 3 The elements of
        int value = 3;
        // Declare end value
        int end = nums.length;
        // use for Cycle is knowing the number of cycles , This is used here while Because I don't know how many times I can cycle , And the start value in the algorithm can only be less than the end value
        while (start < end) {
            // Each cycle redefines the intermediate value
            middle = (start + end) / 2;
            // When the intermediate element is equal to the searched element , Output the index of the intermediate element , return
            if (value == nums[middle]) {
                System.out.println(" The array index of this number is " + middle);
                return;
            } else if (value > nums[middle]) {
                // If the intermediate element is less than the lookup value , Start value plus one
                start = middle + 1;
            } else if (value < nums[middle]) {
                // If the intermediate element is greater than the lookup value , End value minus one
                end = middle - 1;
            }
        }
        // No value was found in the above loop , Then it proves that there is no
        System.out.println(" No such number ");
    }

}