use of static cast

Hey all,
I have been studying this exercise from "C++ Without Fear". I am new at C++ so it took me a while to follow through the whole program. I have a question though..
Why is var is_prime needed to be declared as int? The program used a static cast for double. Can't it be declared as double from the beginning?
It is just a little confusing to me.. Thanks guys and gals, here's the code:

Code:

---

#include <stdafx.h>
#include <iostream>
#include <math.h>
using namespace std;

int main ()
{
        int n;  //number to test for prime-ness
        int i;  //loop counter
        int is_prime;  //Boolean flag

        //Assume that a number is prime until proven otherwise

        is_prime = true;

        //Get a number from the keyboard

        cout << "Enter a number and press ENTER: ";
        cin >> n;

        /*Test for prime-ness by checking for divisibility by all whole numbers from 2 to sqrt (n).*/

        i = 2;
        while (i <= sqrt(static_cast<double>(n))) //While i is <= sqrt(n),
        {
                if (n % i ==0)          //If i divides evenly into n,
                        is_prime = false;    //n is not prime.
                i++;                                        //Add 1 to i.
        }
        //Print results

        if (is_prime)
                cout << "Number is prime." << endl;
        else
                cout << "Number is not prime." << endl;
        return 0;
}

---

Well, the first question, is_prime is declared int, but it really should be declare "bool" in my opinion. It may be a case of the author using an older compiler that didn's support "bool" as a type.

The second question, I take, is why is n declared int, rather than double, and thus forcing the use of static_cast for the sqrt() call... Well, if n is not an integer, you can't really use the modulo operation (and even if you could, it would be quite a bit slower).

Of course, you don't really want to use sqrt() in a condition. It would be better to do something like

Code:

---

  int x = sqrt(static_cast<double>(n));
  while(i < x && is_prime) {
    ...
  }

---

This way, the sqrt() isn't calculated each loop, and the loop ends when we find that a number isn't a prime.

--
Mats

Code:

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while (i <= sqrt(static_cast<double>(n)))

---

That is pretty silly, really. For one thing, passing an int to a function that is expecting a double will result in an implicit cast, as long as you have the prototype for the function, which you do. (sqrt()'s prototype is in <math.h> or <cmath>.)

Code:

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int x = sqrt(static_cast<double>(n));

---

That wouldn't do, either. The return value of sqrt() is a double, and that should be casted before being stuffed into an int. I'd use something like

Code:

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int x = static_cast<int>(sqrt(n));

---

Besides, there's a much simpler way to do this! Why not just use this?

Code:

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while(i*i <= n)

---

Sure, i*i might overflow for large numbers, but n is of the same type as i, so it doesn't matter.

In C++, sqrt() and the other <cmath> functions are overloaded to take and return any of the standard floating-point types (float, double, long double). I'm not sure what it does if passed an integer type.

http://www.cplusplus.com/reference/c...math/sqrt.html

> Besides, there's a much simpler way to do this! Why not just use this?

That's probably safer also, since one has to make sure that if n is an exact square, then i actually takes on the highest value and that rounding error doesn't prevent this. It's probably also faster than sqrt().