hello,i must solve this problem : "Sum all even digits of a given number. That's what i've got so far.

#include <iostream>
using namespace std;
int main() {
int sum = 0 , num;
cin>>num;
while(num > 0) 
{
	sum+=num%10;
	num/=10;
}
cout<<sum;
	return 0;
}

the problem is that this program sum all digits (odd & even), but i need to sum even digits... help pls.

When you add num%10 to sum you need to test if its a even number.A number is even if the number % 2 == 0 meaning that the remainder of the number divided by 2 is 0 since all even number are divisible by 2.

#include <iostream>
using namespace std;
int main() 
{
	int sum = 0 , num=10,cnt=0;
	//cin>>num;
	cout<<"Sum all even digits of a given number  \n\n\n";
	cout<<"Number = "<<num<<"\n\n\n";

	while(cnt < num) 
	{
		cnt++;
		if (cnt % 2 ==0)
		{
			sum+=cnt;
			cout<<cnt<<" + ";
		}
		
	}
	cout<<"  = "<<sum<<" \n\n\n";
		return 0;
}
#include <cstdlib>
#include <iostream>
#include <numeric>
#include <stdio.h>

using namespace std;


int char_add(int sum, char character)
{
   return sum + (character - '0');
}

int main()
{
   cout << "enter number: ";
   string num;
   cin >> num;
   

   cout << "Total: "
        << accumulate(num.begin(), num.end(), 0, char_add)
        << '\n';






    system("PAUSE");
    return EXIT_SUCCESS;
}
This question has already been answered. Start a new discussion instead.