Adding the sum of the digits in an integer

[proudHelios]

Hello, I read that I'm not supposed to ask for the solution to homework assignments, which is fine by me. I'd rather work out the problem myself, else I'll never be able to make it on my own.

However, that being said, I'm having some difficulty with my latest assignment.

We are instructed to use a for loop to add the sum of the digits in an integer. So, 3426 would be 3 + 4 + 2 + 6 = 15. Simple enough, right? I thought so... But now I'm not so sure.

So, what I have done is use the % operator (number%10) to get the value of the last digit and I store it to numbersum (type int).

The problem I am encountering is trying to get it to read the next digit after that. In other words, for 3426, it's reading the 6 just fine, but it won't read the 2, 4, or 3.

As we are not allowed to use any functions, would someone please be so kind as to explain which method or operation I might be able to use to get my program to read the next digits in my integer?

Thank you!

[omegazero]

if it is an int, you could just do something like this...
3426 % 10 gets you the 1's, now divide by 10, its a decimal... the decimal gets dropped off for the next part of the for loop. so it would just repeat.
assuming its only 4 digits, no more no less,

int number
int numbersum = 0
int x = 1
cin >> number
for (x = 1; x <= 4; ++x){
numbersum = numbersum + number % 10
number = number / 10
}

obviously, this wouldn't be finished code, but it would get the results you want.

[cheif]

omegazero was kind enough to give you some potentially confusing code rather than the answer :). The answer is in there though.

What is the result of 3426 / 100? How about 3426 / 1000? How about 3426 / 10000? Understanding that pattern, combined with understanding the % operator, and a loop, should form together into your solution.

[aakashjohari]

This is your solution -

#include <iostream>

using namespace std;

int main()
{
int num;
int sum = 0;

cout << " Enter a number : ";
cin >> num;

while ( num > 0 ) {
sum += num % 10;
num /= 10;
}

cout << "Sum = " << sum;

return 0;
}

[proudHelios]

Thanks everyone! I actually gathered the concept from the first response. It went off in the back of my head kind of like a gong. I suppose that sometimes I just have issues seeing the simpler solutions to things.

[QWERTYman]

Actually, from a certain point of view, you all are wrong. This is impossible.

All operators are functions in disguise.

x + y is actually x.operator+(y), or operator+(x, y), or whatever.

I realize this is unnecessary, but I needed to find someplace to vent my anger. Yay!