Adding odd numbers in a line of input

[daft science]

First time posting. Been lurking for a while though, you guys have helped me out immensely so far though! Thanks :D
My question is about taking a line of input (32677) and adding all the odd numbers in it (3+2+7+7). All the questions similar to that in my search have to do with a range of odd numbers which I know how to do, just not how to take the actual numbers from the input. Hopefully I'm not asking something that's been answered a billion times before. Do I need to use strings to do this? Here's what I have so far but I'm pretty sure I'm missing something pretty basic.

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

/*
* the sum of all odd digits of an input.
* (for example, if the input is 32677,
* the sum would be 3 + 7 + 7 = 17)
*/
int main(int argc, char** argv) {

int num;
int odd =0;

cout << "Please enter a number : ";
cin >> num ;

if (num%2!=0)
{
odd += num;
}

cout << "The sum of all the odd numbers is : " << odd << endl;
return 0;
}

[hamsterman]

With strings (don't forget to #include <string> )

string in;
cin >> in;
int odd = 0;
for(int i; i < in.length(); i++)
   if(in[i] >= '0' && in[i] <= '9'){ //is it a digit?
      int digit = in[i]-'0';//subtracting the ascii value for '0' gives the actual integer
      if(digit %2 != 0) odd += digit;
   }else{
      cout << "what's this!?";//using a string does not ensure that only digits will be entered
      return 0;
   }

Without strings

int n;
cin >> n;//if the user enters a word, this gets stuck, or writes garbage to n, I don't remember
int odd = 0;
while(n > 0) {
   int digit = n % 10;//get the last decimal digit
   n /= 10;//erase the last digit from n;
   if(digit % 2 != 0) odd += digit;
}

[daft science]

Worked beautifully, thanks. :)

I don't completely understand why lines 5 and 6 work.
Line 7 makes sense to me but why do you need an additional number's remainder of 10 and then divide the original number by ten and not use the original number in the if statement? Would you be able to give me a pseudo code example so I can understand the concept?

[Chervil]

Try this, look at the output as well as the code, see if you understand what is going on:

#include <iostream>

using namespace std;

int main()
{

    int n = 314159265;

    while (n > 0)
    {
        cout << "n = " << n;                   // display current value of n
        int digit = n % 10;                    // get the last decimal digit
        n /= 10;                               // erase the last digit from n;
        cout << "  digit = " << digit << endl; // display current digit
    }

    return 0;
}

n = 314159265  digit = 5
n = 31415926  digit = 6
n = 3141592  digit = 2
n = 314159  digit = 9
n = 31415  digit = 5
n = 3141  digit = 1
n = 314  digit = 4
n = 31  digit = 1
n = 3  digit = 3

[AbstractionAnon]

I assume you're referring to the second set of code hamsterman posted.

Some times it helps to work through the steps with pencil and paper.
Let's start with your original number of 32677.
We need to isolate the last digit (7). Line 5 does that for us with the modulo 10 operation.
We'll come back to line 6 in a moment.
Line 7 tells us that the last digit is odd.
Now on the second time through the loop we want to examine the 10's digit, but we don't have an operation that will readily give us the 10's digit. That's were line 6 comes in. Line 6 divides n by 10 leaving the result in n. Now we have 3267 and our second pass through the loop will look at what was the 10's digit which is now in the one's position. Third pass through the loop, n becomes 326 and we're now looking at what was the 100's position in the one's position, etc.

[hamsterman]

I'm not sure what you're asking about. Sure, you can use n in the if, but then it's

if(n % 2) odd += n % 10;
n /= 10;

[daft science]

Thanks for all the help. I'm pretty sure I get it now and all my loop programs are running. :3D Thanks so much you guys.