Why does the program ends even before running?

Question

I want to use this question to improve a bit in my general understanding of how computer works, since I'll probably never have the chance to study in a profound and deep manner. Sorry in advance if the question is silly and not useful in general, but I prefer to learn in this way.

I am learning c++, I found online a code that implements the Newton-Raphson method for finding the root of a function. The code is pretty simple, as you can see from it, at the beginning it asks for the tolerance required, and if I give a "decent" number it works fine. If instead, when it asks for the tolerance I write something like 1e-600, the program break down immediately and the output is Enter starting value x: Failed to converge after 100 iterations .

The output of failed convergence should be a consequence of running the loop for more than 100 iterations, but this isn't the case since the loop doesn't even start. It looks like the program knows already it won't reach that level of tolerance.

Why does this happen? How can the program write that output even if it didn't try the loop for 100 times?

Edit: It seems that everything meaningless (too small numbers, words) I write when it asks for tolerance produces a pnew=0.25 and then the code runs 100 times and fails.

The code is the following:

#include <iostream>
#include <cmath>
using namespace std;
#define N 100    // Maximum number of iterations


int main() {

    double p, pnew;
    double f, dfdx;
    double tol;
    int i;

    cout << "Enter tolerance: ";
    cin >> tol;
    cout << "Enter starting value x: ";
    cin >> pnew;
    // Main Loop
    for(i=0; i < N; i++){
        p = pnew;
        //Evaluate the function and its derivative
        f = 4*p - cos(p);
        dfdx= 4 + sin(p);
        // The Newton-Raphson step
        pnew = p - f/dfdx;
        // Check for convergence and quit if done
        if(abs(p-pnew) < tol){
            cout << "Root is " << pnew << " to within " << tol << "\n";
            return 0;
        }
    }
    // We reach this point only if the iteration failed to converge
    cerr << "Failed to converge after " << N << " iterations.\n";
    return 1;
}

Answer 1

1e-600 is not representable by most implementations of double. std::cin will fail to convert your input to double and fall into a failed state. This means that, unless you clear the error state, any future std::cin also automatically fails without waiting for user input.

From cppreference (since c++17) :

If extraction fails, zero is written to value and failbit is set. If extraction results in the value too large or too small to fit in value, std::numeric_limits<T>::max() or std::numeric_limits<T>::min() is written and failbit flag is set.