Answer to Question #335954 in C++ for GANIZANI

Question

The function below calculates and returns the sum of integers between 1 and n (inclusive) where n is an integer accepted by the function i.e.

If n = 3;

then the function returns 6 (1+2+3). int sum(int n) { int sum = 0;

while (n > 0) { sum = sum + n; --n; } return sum;

}

Rewrite the body of the sum function using a single do-while loop instead of a while-loop

Answer 1

#include <iostream>
using namespace std;

int sumN(int);

int main() {
 int n;
 
 cout<<"N: ";
 cin>>n;
 if(n>0){
     cout<<sumN(n)<<endl;
 }
 
  return 0;
}

int sumN(int n){
    int sum=0;
    
    do{
     sum = sum + n;
     --n;    
        
    }while (n > 0);
    
     return sum;
}