Notan Deward - 3 months ago 7
C++ Question

# Sequence of integers using While loop

Hello Everyone i am new to C++ Programming a College Freshmen.
(this not a homework or project, he just give me some task to do)
i was given a task to write and run a program that reads a sequence of integers until a negative integer is entered, then prints the sum of the positive even and positive odd integers. Use a While Loop.

Basically the user have infinite amount to enter integers and it will only stop if he entered a negative integers and it will display all the number of Even and Odd.

i try my best to do it by myself without asking for help in the internet but this what i got. i am still new to programming and i'm looking forward for you help. better if you can complete it and i will study out the codes.

``````#include <iostream>
using namespace std;

int main()
{
int n;
int odds =0;
int evens =0;
cout<<"Enter Positive integers.  \n\t:";
cin>>n;
while (n>0)
{
switch(n%2)
{
case 0:
evens++;
break;
case 1:
odds++;
}

}
cout<<endl;

cout<<"the number of odds: "<<odds<<endl;
cout<<"the number of evens: "<<evens<<endl;

}
``````

This is what you need:

``````int main()
{
int n;
int odds = 0;
int evens = 0;

while(cin >> n) {
if(n < 0) {
break;
}
else {
if((n % 2) == 0) {
++odds;
}
else {
++evens;
}
}
}
cout  << "the number of odds: "<< odds << endl;
cout  << "the number of evens: "<< evens << endl;
}
``````

I wrote this code as simple as possible so you can understand what is going on. The `while(cin >> n)` keeps reading values(it will terminate if an `EOF` flag is set). Then the program checks if the number is negative and exits the for loop, otherwise it calculates if the number is `odd` or `even`. You can experiment with this code by adding switch statements instead of ifs.

Source (Stackoverflow)