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Thread: Having a problem with calling the sqrt function trying to pass an integer

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  1. #1
    Join Date
    Sep 2019

    Having a problem with calling the sqrt function trying to pass an integer

    I'm having a problem calling the sqrt function in my prime number checker program. My intention is to make the for loop only check up to the square root of the number the user has entered. But C++ Builder is telling me there is ambiguity between sqrt(double) and sqrt(long double).

    I want to pass an integer to the sqrt function but I don't know if I can. Cast it to an int perhaps?

    Here is what I got:

    #include <iostream>
    #include <cmath>
    #include <conio.h>
    using namespace std;
    bool isPrime(int num);
    int main()
    	int x;
    	bool flag;
    	cout << "Enter a positive integer: ";
    	cin >> x;
    	flag = isPrime(x);
    	if (flag == true)
    		cout << x << " is a prime number" << endl;
    		cout << x << " is not a prime number" << endl;
    	return 0;
    bool isPrime(int num)
    	bool flag;
    	for (int i = 2; i <= sqrt(num); i++)
    		if(num % i == 0)
    			flag = false;
    	return flag;
    Last edited by 2kaud; 09-10-2019 at 04:17 AM. Reason: Added code tags

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