Sum of digits of some number

Question

for example, if I typed in 12345 then sum = 1+2+3+4+5 Using C I used some kind of approach that I'm not sure about it and there is a problem with the line referred to down in the code

int i ;
int sum;
int individual;
int n;
printf("enter the number: ");
scanf("%d",&n);
for (i=0;i<5;++i){
  **indvidual=+n[i];**
  sum=+invidual;
}
printf("%d",sum)
return 0;

Answer 1

You can divide by 10 and see the remainder to obtain each digits.

Also you should use += , not =+ , to add things to variable and initialize the variables before using their values.

int i ;
int sum = 0; /* initialize */
int individual;
int n;
printf("enter the number: ");
scanf("%d",&n);
for (i=0;i<5;++i){
  indvidual=+(n%10); /* obtain a digit (+ is not required, but left by respect) */
  n/=10; /* eliminate last dight and proceed to next digit */
  sum+=invidual; /* use += instead of =+ */
}
printf("%d",sum); /* also add semicolon here */
return 0;

Answer 2

#include <stdlib.h>    
#include <stdio.h>
int main()
{
    int i = 12345;
    int sum;
          
    i = abs(i);
    sum = i?i%9?i%9:9:0;

    printf ("Sum of digits is %d\n",  sum);
}

Result is 6

Because 1 + 2 + 3 + 4 + 5 = 15 and then 1 + 5 = 6

(or is the expected output 15 , after just a single pass through the digits?)

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If you want only a single pass through the digits:

 #include <stdlib.h> #include <stdio.h> int main() { int i = 12345; int sum=0; i = abs(i); while(i) { sum += i%10; i /= 10; } printf ("Sum of digits is %d\n", sum); return 0; }