Putting an IF statement inside a Loop? - c++

Question

I am trying to write a program that accepts inputs of even numbers until the user enters a odd number then it stops and adds all the even inputs.
The problem I am having is that I'm intending to use an if statement to determine if the number is even or not but once the program has run I need it to run again so the user can keep inputting numbers until they input a odd one.
Here's what I have so far:
#include "stdafx.h"
#include <iostream>
using namespace std;
int main()
{
int value;
cout << "Please enter a possitive number: ";
cin >> value;
if (value%2 == 0) { // divides the value entered by 2 too determine whether or not its even.
cout << "Please enter another even number: ";
}
else
{
}
cin.get();
return 0;
}

Answer 1

You need a loop in order to ask for input again in case the value is even.
You will enter the while loop only if value is even. And you will loop until value input is even.
You then need another int which represents the sum and add the value to the sum each time you enter the loop (each time you have an even number)
Note that with this solution you don't need an if statement. The condition is in the while loop
#include <iostream>
using namespace std;
int main()
{
int value;
int sum=0;
cout << "Please enter a possitive number: "<<endl;
cin >> value;
while (cin && value%2 ==0){
sum+=value;
cout << "Please enter another even number: "<<endl;;
cin >> value;
}
cout<<"Sum of even number: "<<sum<<endl;
return 0;
}

Answer 2

Just do it.
#include <iostream>
#include <vector>
using namespace std;
int main()
{
int value;
vector<int> inputs;
cout << "Please enter a possitive number: ";
while (cin >> value) { // loop while successfully read an integer
if (value%2 == 0) { // divides the value entered by 2 to determine whether or not its even.
inputs.push_back(value);
cout << "Please enter another even number: ";
}
else
{
break; // get out of this loop
}
}
// add all the even inputs
int sum = 0;
for (vector<int>::iterator it = inputs.begin(); it != inputs.end(); it++) {
sum += *it;
}
cout << sum << endl;
return 0;
}
Alternatively, you can do the addition inside the loop.
#include <iostream>
using namespace std;
int main()
{
int value;
int sum = 0;
cout << "Please enter a possitive number: ";
while (cin >> value) { // loop while successfully read an integer
if (value%2 == 0) { // divides the value entered by 2 to determine whether or not its even.
sum += value;
cout << "Please enter another even number: ";
}
else
{
break; // get out of this loop
}
}
cout << sum << endl;
return 0;
}

Answer 3

something like this?
int value;
bool run = true;
cout << "Please enter a possitive number: ";
while(run)
{
cin >> value;
if (value%2 == 0) { // divides the value entered by 2 too determine whether or not its even.
cout << "Please enter another even number: ";
}
else
{
run=false
}
}
It is sudo code modeled similar to your loop, so not the best way, also most likely there might be syntax error, but should give you an idea

Answer 4

one more way to check whether it is even or odd is to check lowest bit:
#include <vector>
#include <iostream>
#include <algorithm>
int main()
{
int enteredValue = 0;
std::vector<int> result;
do {
std::cout<<"Please enter even number : ";
std::cin>>enteredValue;
result.push_back(enteredValue);
std::cout<<'\n';
}while(!(enteredValue & 0x01)); //check the lowest bit
std::for_each(result.begin(), result.end(), [](int x){ std::cout<<x<<' '; });
return 1;
}