Can someone test this code and see why it's not working? Your supposed to enter a number, and it tells you if it's prime or not.

Question

#include <iostream>
using namespace std;

int main(){
int div=2;
int num;
bool sw = true;
cout<<"Please enter a number:"<<endl;
cin>>num;
do{
if(num%div>0){
++div;
}
else{
sw=false;
}
}while(div<num&&num%div>0);
if(sw==true){
cout<<num<<" is a prime number.";
}
else{
cout<<num<<" is not a prime number.";
}
return 0;
}

Answer 1

why can't you go with for loop ?

one more thing is that one need to check for half of the input number... for example, for 10 as input, modulo need to be check for 5 only to decide 10 as prime number...

Half of number is less efficient compared to square root of number... check till 3 only for 10 as input and you are done..

Answer 2

You shouldn’t use do...while. Try inputing 2 and seeing why that doesn’t work.

Answer 3

The more important bug is that suppose num is 9, and div is 2. num % div is not 0, so increment div. Now div is 3, and the while condition finds that num % div is 0, so it breaks out of the loop! We have not set sw to false yet, so the program says it’s prime.

Answer 4

Please fix your coding style, it is what caused your error. Here's a cleaned and working version.

#include <iostream>
using namespace std;

int main() {
int div=2;
int num;
bool sw = true;

cout << "Please enter a number: " << endl;
cin >> num;

do {
if(num % div > 0)
++div;
else
sw = false;
} while(div < num && num % div > 0);

if (sw == true)
cout << num <<" is a prime number.\n";
else
cout << num << " is not a prime number.\n";

return 0;
}

Answer 5

So, what should I use instead of a do while loop?

Answer 6

Just a while loop. Do while always enters the loop at least once, but you should not enter the loop when num == 2