Thread: Sum of digits of an integer: Odd or Even?

Hey, I've been trying to work out a function for this for a good while, I'm brand new to loops so I'm not sure which would be the best to use to separate the digits out to add. I do know from some searching around that I need to do:

number % 10 to get the digit alone

number / 10 to strip away the right-most number, I'm just having troubles with looping it.

Thanks in advance!

I'm sure any type of loop will do, it doesn't matter.

>> number / 10 to strip away the right-most number, I'm just having troubles with looping it.
Think about what that means, though. All positive numbers have a lower bound of zero, so when are you done?

Would you be done when number < 1?

If number is always positive, yes.

Could I do something like:

Code:

{
	int number = 0, digit = 0;

	for(number != 0)
	{
		digit = number % 10;

		number = number / 10;
	}

}

Well, I got the loop figured out, but now I'm having trouble adding the digits of the integer together (eg. 1987; 1+9+8+7). I don't really know how to store or save the digits after they have been singled out. Any help is much appreciated!

When you get the first digit, you would add it to the sum of zero. And then start adding more digits until you run out.

sum = sum + digit

Ok, thanks for the help! Would this function do the trick?

Code:

int isOddEven(int intNumber)
{
	int number = 0, digit = 0, sum = 0;

	while(number != 0)
	{
		digit = number % 10;

		sum = sum + digit

		number = number / 10;
	}

	return sum;

}