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

  1. #1
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    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!

  2. #2
    Lurking whiteflags's Avatar
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    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?

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    Would you be done when number < 1?

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    Lurking whiteflags's Avatar
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    If number is always positive, yes.

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    Could I do something like:
    Code:
    {
    	int number = 0, digit = 0;
    
    	for(number != 0)
    	{
    		digit = number % 10;
    
    		number = number / 10;
    	}
    
    }

  6. #6
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    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!

  7. #7
    Lurking whiteflags's Avatar
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    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

  8. #8
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    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;
    
    }

  9. #9
    and the Hat of Guessing tabstop's Avatar
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    Quote Originally Posted by Devolution View Post
    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;
    
    }
    Apart from a missing ";" and the fact that the name of the function is "isOddEven" but returns the sum of the digits, and the fact that you will always return 0 since you're using the local variable number instead of the parameter passed into the function.

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    Oops, yea I got the function name switched up with another one.

    How do I return the parameter passed into the function? I thought returning sum would do that, but I'm pretty new to C, and things haven't "clicked" for me yet I'm trying my best to understand so I don't fall behind.

  11. #11
    and the Hat of Guessing tabstop's Avatar
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    Quote Originally Posted by Devolution View Post
    Oops, yea I got the function name switched up with another one.

    How do I return the parameter passed into the function? I thought returning sum would do that, but I'm pretty new to C, and things haven't "clicked" for me yet I'm trying my best to understand so I don't fall behind.
    You don't want to return the parameter passed into the function, and I didn't suggest you do so. I suggest you use the parameter passed into the function, instead of completely ignoring it in favor of calculating the sum of the digits of 0 (a/k/a number) instead.

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    Ok I believe I understand what you're saying now, correct me if I'm wrong though. The "number" variable in there too was another mix-up that I didn't mean to have in the function, but thanks for pointing it out. Fixed up, it looks like this:

    Code:
    int sumDigits(int intNumber)
    {
    	int digit = 0, sum = 0;
    
    	while(intNumber != 0)
    	{
    		digit = intNumber % 10;
    
    		sum = sum + digit;
    
    		intNumber = intNumber / 10;
    	}
    	return sum;
    }

  13. #13
    and the hat of sweating
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    I don't really see the point of your function.
    To check if a number is even or odd, you just use the % operator and specify a certain number afterwards, and that number isn't 10.
    "I am probably the laziest programmer on the planet, a fact with which anyone who has ever seen my code will agree." - esbo, 11/15/2008

    "the internet is a scary place to be thats why i dont use it much." - billet, 03/17/2010

  14. #14
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    We're checking to see if the sum of the digits of an integer are even or odd.

  15. #15
    Banned ಠ_ಠ's Avatar
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    Quote Originally Posted by Devolution View Post
    Ok I believe I understand what you're saying now, correct me if I'm wrong though. The "number" variable in there too was another mix-up that I didn't mean to have in the function, but thanks for pointing it out. Fixed up, it looks like this:

    Code:
    int sumDigits(int intNumber)
    {
    	int digit = 0, sum = 0;
    
    	while(intNumber != 0)
    	{
    		digit = intNumber % 10;
    
    		sum = sum + digit;
    
    		intNumber = intNumber / 10;
    	}
    	return sum;
    }
    instead of "sum = sum + digit;" you can just write "sum += digit;"
    same thing for intNumber "intNumber /= 10;"

    saves a little bit of typing
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