adding odd numbers

hi
if i have a list of numbers , say 2345678
and want to add up the odd position digits
ie 2+4+6+8

how can i do this ?

i tried it with nums[20]
and using atoi to convert the type

but it seems i can't do things like

sum=atoi(nums[0]) + atoi(nums[1]);

can someone plz help me~~~~

Try this:

PHP Code:

---

char nums[]="2345678"
int i;
int add=0;
for(i=0; i < strlen(nums); i + 2)
{
add = add + (atoi(nums[i]) );   /// OR add += (atoi(nums[i]) )
}
printf("%i", add); 


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try changing :-

for(i=0; i < strlen(nums); i + 2)
to
for(i=0; i < strlen(nums); i += 2)

this will do it for any number

Code:

---


#include <stdlib.h>
#include <string.h>

#define MAX_SIZE 15


int function(int num)
{
    int ret = 0;
    char number[MAX_SIZE];
    memset(number,'\0',MAX_SIZE);
    _itoa(num,number,10);
    for(int i = 0;number[i] != '\0';i++)
  {
        if((i % 2 != 0) || i == 1)
            ret += i;
  }
  return ret;
}

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Does it have to be a string

int num[7] ={2,3,4,5,6,7,8};

then

for(i =0; i < 7;i++)
if(i % 2 == 0)
total += num[i];

should do the trick

A third way.

Code:

---

#include <stdlib.h>
#include <stdio.h>

int add_odd_digits(int number);
int main()
{
  int num = 12345;
  int sum;
  sum = add_odd_digits(num);
  printf("sum of odd digits:%d\n",sum);
  return 0;
}

int add_odd_digits(int number)
{
  int digit, odd_sum;
  int sum1 = 0;
  int sum2 = 0;
  while (number > 0)
  {
      digit = number % 10;
      number /= 10;
      sum1 += digit;
      if (number <=0)
        return sum1;
      digit = number % 10;
      number /= 10;
      sum2 += digit;
      if (number <=0)
        return sum2;
  }
}

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