C Program to find sum of all digits in number is simple program.
In this tutorial, we are going to implement a new program which is to calculate the sum of all digits in number.
sum of all digits in number
First, we have a number given any number of digits in it. We need to add every digit present in number.
How to sum of all digits in a number?
Steps:-
To do this first we have to separate digits present in number.
For separating digits we can use modulus operation (% 10)which gives us a single digit.
After getting single digit we perform sum operation.
We Repeat step 1 and 2 till digits in number end.
Example
Suppose the number = 365
1. Now to perform sum of all digits we have to first separate out each digital. We do this by modulus which gives us reminder ( a digit)
=> rem = number % 10
=> rem = 365 % 10
=> rem = 5
2. Add rem (a digit)
=> sum = sum + rem
=> sum = 5
3. Repeat step 1 and 2
=> rem = number % 10
=> rem = 36 % 10
=> rem = 6
4. Add rem (a digit)
=> sum = sum + rem
=> sum = 5 + 6
=> sum = 11
5. Repeated 1 and 2
=> rem = number % 10
=> rem = 3 % 10
=> rem = 3
6. Add rem (a digit)
=> sum = sum + rem
=> sum = 11 + 3
=> sum = 14
C Program to find sum of all digits in number
| 1234567891011121314151617181920212223242526272829 | #include<stdio.h>main(){int dummy,n,sum=0,x;printf(“Enter a number\n”);scanf(“%d”,&n);dummy=n;while(n>0){ x=n%10; sum=sum+x; n=n/10;}printf(“The sum of all digits in %d is %d\n”,dummy,sum);} |
Output
Explanation
1. Here we did initialization for
dummy ->To store the entered value(i.e ‘n’) as you will come to know at the end of the program
n –>To store number given by a user.
sum->To store the sum of all digits in the number.It is initialized to zero
x->To store n%10.
2. First of all, we got a number ‘n’ from a user and then stored it in a dummy variable called as ‘dummy’ for restoring the value.(remember this point)
3. Now the main logic comes here
let the number ‘n’ be 321 and as 321>0, while loop gets executed
then x=321%10—>which is 1.
sum=0+1——–>1
n=321/10———>32
The sum for the first loop execution is sum=1.
4. Now the number ‘n’ has become ’32’ and n>0, while loop executes for the 2nd time
then x=32%10—>which is 2.
sum=1+2——–>3
n=32/10———>3
The sum when loop executed the second time is sum=3.
Now the number ‘n’ has become ‘3’ and n>0, while loop executes for the 3rd time
then x=3%10—>which is 3.
sum=3+3——–>6
n=3/10———>0
The sum when loop executed the third time is sum=6.
Now as the number of variable ‘n’ is 0 which is not n>0 then the loop terminates.Then the final sum is ‘6’.
So now I hope you understood why the dummy variable is used.It is because the value in ‘n’ becomes 0 at the end of the program so for restoring this value to print at the end we used ‘dummy'(as from the 2nd point).
5. Finally, it prints the value in ‘sum’.