use of static cast

This is a discussion on use of static cast within the C++ Programming forums, part of the General Programming Boards category; Hey all, I have been studying this exercise from "C++ Without Fear". I am new at C++ so it took ...

  1. #1
    Registered User
    Join Date
    Dec 2006
    Jacksonville, AR

    Smile 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:

    #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;
    		cout << "Number is not prime." << endl;
    	return 0;

  2. #2
    Kernel hacker
    Join Date
    Jul 2007
    Farncombe, Surrey, England
    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
      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.

    Compilers can produce warnings - make the compiler programmers happy: Use them!
    Please don't PM me for help - and no, I don't do help over instant messengers.

  3. #3
    Frequently Quite Prolix dwks's Avatar
    Join Date
    Apr 2005
    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>.)

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

    Seek and ye shall find. quaere et invenies.

    "Simplicity does not precede complexity, but follows it." -- Alan Perlis
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  4. #4
    Registered User
    Join Date
    Sep 2006
    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.

    > 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().
    Last edited by robatino; 10-10-2007 at 09:11 PM.

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