使用atoi将字符串索引转换为数组索引

Question

I'm doing a code which calculates the sum of the digits of a factorial, my solution was to convert the number to a string then put that string on an array.

I tried using atoi to convert a string index to the array index, but it doesn't work, giving me the error "passing argument 1 of 'atoi' makes pointer from integer without a cast"

#include <string.h>
#include <stdlib.h>
int fat(int x)
{
    if (x == 0 || x == 1)
    {
        return 1;
    }
    else
    {
        return x * fat(x-1);
    }
}

int main()
{
    int n, i, f=0, arr[30];
    char str[30];
    printf("Type the value of N: ");
    scanf("%d",&n);
    for (i=1;i<=n;i++)
    {
        f = fat(i);
    }
    printf("%d \n", f);
    sprintf(str, "%d", f);
    n=0;
    for (i=0;i<strlen(str);i++)
    {
        arr[i]=atoi(str[i]);
        n=n+arr[i];
    }
    printf("%d", n);
}

Answer 1

If you just want to convert a single digit character to a number, you don't need to use atoi() . Use str[i] - '0' .

        arr[i] = str[i] - '0';

There also doesn't seem to be much point to the array. You can just do:

        n += str[i] - '0';

without saving all the numbers in an array that you never use again.