how to sum the digits of a number?

Going fine so far. Now you need to make use of recursion passing sumb.

Of course usually everything adds up to 23 :p and you start losing your mind...

(for those that don't get it, it was a reference to a certain Jim Carey movie)
Tis true though - Both my and my partners date of birth add up to 23 :eek:

i cant use recursion only loop
what to do?

There is no sense doing digit = something and digits = something else. You always get something else. Delete the red line.

Code:

---

printf("enter digit");
scanf("%d",&digit);
do
{
sumb += digit % 10;
digit=digit-digit%10;
digit = digit / 10;
}while(digit / 10 >1);
printf("%d",sumb);

---

So you get the first part. The sum of all digits. Now you need to do the same method with sumb. So, make a function with your algorithm in which you will pass digits. Then use a loop with the function.

Code:

---

int sumDigits(int digit) {... return sumb}
int main()
{
  int result = 0;
  ...
    for (...) {
        ...
        result = sumDigits(...)
    }
    printf("Sum =%d", 23);
}

---

I cant use recursion or function call.

i tried to built this one
but for 679 it gives me 7
i dont know why?

Code:

---


 printf(" enter number  ");

scanf("%d",&digit);
copy = digit;


do
{
sumb=sumb+copy%10;

copy=copy/10;

if (sumb<10){
    flag=1;
}

else
{
copy=sumb;
sumb=0;
}

}while((copy!=0)&&(flag=1));

---

You set flag=1 the first time through, and it never gets cleared when the sum goes over 10.

no
i declared
int flag=0;
and i put flag=1 only if the sum under 10

where is the mistake

Quote:

---

Originally Posted by transgalactic2

I cant use recursion or function call.

---

My hint would be to use two variables and two nested (while) loops. One variable would be used to store the original input while the other would be a temporary to store the current sum. The outer loop would be used to keep looping until the sum is less than 10. The inner loop would be used to compute the current sum.

what about my code
how to fix it
?

Please indent your code. You should also post the smallest and simplest program that demonstrates the error.

One possible problem is that you write (flag=1) when you probably mean (flag == 1). However, it would be good if you stated what is the algorithm that you are trying to implement.

is it intended now?
i fixed the "==" thing
its still doesnt work
??

Code:

---

printf(" enter number  ");

scanf("%d",&digit);
copy = digit;


do
{
sumb=sumb+copy%10;
copy=copy/10;
if (sumb<10){
    flag=1;
}

else
{
    copy=sumb;
    sumb=0;
}
}while((copy!=0)&&(flag==1));

---

Is it so hard to indent the code?

Code:

---

printf(" enter number  ");

scanf("%d", &digit);
copy = digit;

do
{
    sumb = sumb + copy % 10;

    copy = copy / 10;

    if (sumb < 10)
    {
        flag = 1;
    }
    else
    {
        copy = sumb;
        sumb = 0;
    }

} while ((copy != 0) && (flag == 1));

---

Considering that you have printed the wrong result before, I strongly suggest that you post the smallest and simplest compilable program that demonstrates the error (and state what exactly is the error).

As for the fix: I think you actually should write flag != 1 instead, but that's the problem when you use such a generic name as "flag". We do not know what exactly is the intended meaning of the flag being set.

So trace through the !@#$^@#$% thing already. Suppose copy = 679. I assume you managed to set sumb = 0 before we start.
First time through: sumb = 9; copy is 67; flag is set to 1 since sumb is less than 10.
Second time through: sumb = 16; copy is 6; copy is set to 16 and sumb is set back to 0 -- even through we're not done with the number yet! -- and flag is still 1!
Third time through: sumb = 6; copy is 1; flag is set to 1 since sumb is less than 10.
Fourth time through: sumb = 7; copy is 0; flag is set to 1 once again, and now the while loop stops.

i need it to be 679-> 2
i cant see where the mistake in this algorithm
??

Quote:

---

Originally Posted by transgalactic2

i need it to be 679-> 2

---

How do you get that? The recursive sum of the digits of 679 is 4.

Quote:

---

Originally Posted by transgalactic2

i cant see where the mistake in this code

---

As tabstop has pointed out, trace your code. Personally, I feel that you should go with what I suggested. In my opinion, it is easier to implement it using nested loops than trying to figure out the exact conditions for a single loop.

Dude! You have this code:

Code:

---

do
{
    sumb += digit % 10;
    digit = digit / 10;
}while(digit / 10 >1);

---

And you get the sum of all digits.
You want to do this again for the new sumb. So you want to do digit = sumb and do this again.
Like:
digit = 543
sumb = 543 = 5 + 4 + 3 = 12

digit = sumb = 12
sumb = 1 + 2 = 3

Try that in a double while loop. In the end print sumb