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I created a recursive function that uses binary search to just return true if it finds the value and false if it does not.

I'm new to recursion and binary search so please let me know where I can improve upon.

/**
* Uses binary search O(log n). Returns true if the values is in the value array false if it's not. Recursive
*/
bool binarySearch( int value, int values[], int left, int right ) {

    //keep track of when the array will be empty and return false
    if ( right < left ) {
        return false;
    }

    //Find the middle of the array
    int mid = ( left + right ) / 2;

    //compare the value to the middle of the array. If it's a match return true, else move the left position and right position accordingly
    if( value == values[ mid ] ) {
        return true;
    } else if ( value < values[ mid ] ) {
        right = mid - 1;
    } else {
        left = mid + 1;
    }

    //return the function 
    return binarySearch( value, values, left, right );
}
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You return a boolean to indicate whether the value was found or not. The function would be more useful if you returned an index at which the value was found (or -1 if not found). It's better information for the same amount of work.

It appears that left is the leftmost index to consider, and right is the rightmost index to consider. It would be more idiomatic to follow the convention of inclusive-exclusive bounds, with left being the leftmost index and right being just beyond the last element. That way, right - left indicates the number of elements in the array.

You've implemented the search using recursion, but it could also be done using just a loop. You would avoid the overhead of function calls and the possibility of stack overflow.

int mid = ( left + right ) / 2 is vulnerable to integer overflow if both indices are large. int mid = left + (right - left) / 2 would not overflow.

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  • \$\begingroup\$ Awesome, thanks for the advice. So on the point of recursion. Is this not a good case to use recursion? I had originally had it in a loop. \$\endgroup\$ – Joseph Palomino Feb 27 '17 at 23:53
  • \$\begingroup\$ If it can be done in a loop, then generally it should be done in a loop. \$\endgroup\$ – 200_success Feb 27 '17 at 23:54
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    \$\begingroup\$ @200_success: If it can be done in a loop without adding a lot of complexity. Any recursion can be converted to a loop, but in some cases doing so adds a great deal of complexity. \$\endgroup\$ – Jerry Coffin Feb 28 '17 at 0:20
  • \$\begingroup\$ @JerryCoffin actually that's not strictly true... the recursion that can be converted to loops is called "primitive recursion" ... There's also "higher order" recursion. That's recursive programs that cannot be expressed in loops. The prime example I know of that is the Ackermann-Function \$\endgroup\$ – Vogel612 Feb 28 '17 at 0:28
  • \$\begingroup\$ @Vogel612 One can transform any recursion into loop, but in some cases a stack is required. Anyway, the simplicity of bsearch permits one to use recursion with no fear about overhead. \$\endgroup\$ – Gabriel Feb 28 '17 at 1:52

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