
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 primeness
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 primeness 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:
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:
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:
int x = static_cast<int>(sqrt(n));
Besides, there's a much simpler way to do this! Why not just use this?
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 floatingpoint 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().