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The code below is used to perform binary search on a sorted array, then return how many compares it took to find the user-input values. 

int binarySearch(int arr[], int numelems, int value)
    {
        int first = 0, last = numelems - 1, middle, position = -1;
        bool found = false;
        int count = 0;
        while (!found && first <= last)
        {

            middle = (first + last) / 2;
            if (arr[middle] == value)
            {
                found = true;
                position = middle;
                count++;
            }
            else if (arr[middle] > value)
            {
                last = middle - 1;
                count++;
            }
            else
            {
                first = middle + 1;
                count++;
            }
        }
        return count;
    }

Below is the result I get when I call the function to search the number "38". (The result is correct).

enter image description here

I am trying to edit this function so it can also print out the number of compares even if the user-input number does not exist in the array.

So, ideally, if I try to search for the number “24”, the program should print out something like this:

The value 24 does not exist in the array.
It took 5 compares to reach the conclusion.

Somehow I can't quite figure out how to do it... I tried to add an if-statement outside of the while loop, like below 

int binarySearch(int arr[], int numelems, int value)
{
    int first = 0, last = numelems - 1, middle, position = -1;
    bool found = false;
    int count = 0;
    while (!found && first <= last)
    {
        middle = (first + last) / 2;
        if (arr[middle] == value)
        {
            found = true;
            position = middle;
            count++;
        }
        else if (arr[middle] > value)
        {
            last = middle - 1;
            count++;
        }
        else if (arr[middle] < value)
        {
            first = middle + 1;
            count++;
        }
    }
    if (found = false)
    {
        cout << "Value not found.";
    }
    return count;
}

I am not sure how to print out the count even the program didn't find the number, so I simply wrote a cout statement "Value not found." for trial, but even that's not working. If I run the code, this is the result I get

enter image description here

Beans
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    Please post all your code as plaintext. We can't see images, and can't copy code from them. – cigien Apr 12 '20 at 21:42
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    You either generate the output in your search function, or you have your function return two values (how many iterations, and whether it found the value). The answers to [this](https://stackoverflow.com/questions/321068/returning-multiple-values-from-a-c-function) post explain several ways to return multiple values from a function. – JohnFilleau Apr 12 '20 at 21:46

1 Answers1

2

You could simply add an if statement before return, like:

    if (!found)
    {
        cout << "The value " << value << " not found.\n";
    }

Or you could return two values with std::pair (whether the value is found and the count). Based on that, you could print the result at the call site.

P.P
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