Question

Rewrite the function so the formal parameter x is an IO pointer-to int parameter. The function’s prototype is changed as shown below. Please explain code with comments

//--------------------------------------------------

void NextPrime(int *x)

//--------------------------------------------------

{

bool IsPrime(const int x);

}

Answer 1

//C++ program

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

bool IsPrime(const int x){
   //function to check whether a number given in argument is prime or not
   //if any number >=2 and less than equal to its square root not divides x then x is prime
for(int i=2;i<=sqrt(x);i++){
    if(x%i == 0)return false;
   }
   return true;
}

void NextPrime(int *x){
   //increasing value of x by 1 to check whether next element is prime or not
   *x = *x+1;
   //we keep increasing value x by 1 until we find x is a prime
   while(IsPrime(*x) == false){
       *x = *x + 1;
   }
}

//Driver main function
int main(){
   int x = 13;
  
   NextPrime(&x);
  
   cout<<"Next Prime Number : "<<x;
   return 0;
}

//sample output