How can I specifically take out numbers in an integer, in C? - c

Question

Lets say I have an integer called SIN and the scanf input receives 193456787.
so SIN = 193456787;
What I want to do is add up all the other numbers after the first digit.
So 9 + 4 + 6 + 8 = 27
Can somebody please explain to a beginner how to do this?

Answer 1

Print the number and then sum every other digit
int sum_every_other_digit_after_first(unsigned long long x) {
char buf[sizeof x * CHAR_BIT];
sprintf(buf, "%llu", x);
char *p = buf;
int sum = 0;
while (*p) {
p++; // Skip digit
if (*p) {
sum += *p++ - '0';
}
}
return sum;
}
or as inspired by #PageNotFound
int sum_every_other_digit_after_first(unsigned long long x) {
int esum = 0;
int osum = 0;
while (x > 0) {
esum += x%10;
x /= 10;
if (x == 0) {
return osum;
}
osum += x%10;
x /= 10;
}
return esum;
}
or for fun, a recursive solution
int sum_every_other_digit_after_first_r(unsigned long long x, int esum, int osum) {
if (x >= 100) {
int digit2 = x % 100;
esum += digit2 % 10;
osum += digit2 / 10
return sum_every_other_digit_after_first_r(x / 100, esum, osum);
}
if (x >= 10) {
return esum + x % 10;
}
return osum;
}
sum_every_other_digit_after_first_r(1234567,0,0) --> 12

Answer 2

My solution
#include <stdio.h>
int main()
{
int SIN = 193456787;
int a = 0, b = 0, cnt = 0;
while (SIN > 0) {
if (cnt % 2) b += SIN % 10;
else a += SIN % 10;
cnt++;
SIN /= 10;
}
printf("%d\n", cnt%2 ? b : a);
return 0;
}

Answer 3

Note: Please comment if this is not what you intended, as your question is a little ambigous.
#include <stdio.h>
int main() {
unsigned number;
scanf("%u\n", &number);
unsigned result = 0;
unsigned tmp = number;
unsigned numberOfDigits = 0;
do
numberOfDigits++;
while((tmp /= 10) != 0);
if(numberOfDigits % 2 != 0)
number /= 10;
while(number >= 10) {
result += number % 10;
number /= 100; // Skip two digits
}
printf("%u\n", result);
}