Bisect algorithm

#include <iostream>
#include <cmath>

using namespace std;

// This function if is 0 is the solutioon of sqrt(2) - n = 0
double f(double n){
    return sqrt(2) - n;
}

// function that returns the "sign" of the value -1, 0 for 0 or, 1 for positive
template <typename T> int sgn(T val) {
    return (T(0) < val) - (val < T(0));
}

int main()
{
    //------------------ user inputs --------------------------------------

    // define the search interval
    double lower = 0, upper = 2.0;

    // define the tolerance
    double tolerance = 0.001;

    // define the max iterations
    int n_max = 15;


    //------------------ algorithm starts --------------------------------------

    // if the algorithm is precise enough this will be true
    bool result_found = false;



    // start by searching in the middle of the interval
    double half = (upper + lower) / 2.0;

    int n_itr = 0;
    while(n_itr < n_max){
        // calculate the half point
        double res = f(half);

        // if the half point is 0 or the half point is lower than the tolerance
        // end the porgram
        if(res == 0 || ((upper - lower) / 2) < tolerance ){
            cout << "result: " << half << endl;
            cout << "in n iterations: " << n_itr << endl;
            result_found = true;
            break;
        }

        // check which sign is the f(c) and f(lower), if they have the same sign
        // ex. f(c) > 0 and f(lower) > 0, then set the lower limit to the half
        // else the uppper limit is set to half
        // (alternative sign check res * f(lower) > 0)
        if( sgn<double>(res) == sgn<double>(f(lower)) ){
            lower = half;
            half = (lower + upper) / 2;
        }
        else{
            upper = half;
            half = (lower + upper) / 2;
        }

        // shows the interval getting smaller
        cout << "lower: " << lower << " half: " << half <<  " upper: " << upper  << endl;

        n_itr += 1;
    }

    if(!result_found){
        cout << "Not enough iteration to reach the tolerance" << endl;
    }

    return 0;
}