I need to write a function that subtracts from right to left digits of array elements and prints absolute value of them.

For array a[4]={101, 3002, 70008, 900001}, program would print:

OUTPUT: 11 302 7008 90001

Explanation: abs(1-0)=1, abs(0-1)=1, so first printed number is 11

#include <stdio.h>
#include <stdlib.h>
int subtract(int a[], int y, int k) {
    int count[100] = {0}, n[100], b[100][100], number[100], x;
    n[k] = a[k];
    while (n[k] != 0)
    {
        n[k] /= 10;
        count[k]  ;
    }
    number[k] = count[k];
    if (count[k] != 0) {
        count[k] = 0;
        n[k] = a[k];
        while (n[k] != 0) {
            b[k][count[k]] = n[k] % 10;
            n[k] /= 10;
            count[k]  ;
        }
    }
    int i;
    for (i = number[k] - 2; i >= 0; i--) {
        return abs(b[k][i] - b[k][i   1]);
    }
}
int main() {
    int a[4] = {101, 3002, 70008, 900001}, n = 4, i;
    for (i = 0; i < n; i  ) {
        printf("%d ", subtract(a, n, i));
    }
    return 0;
}

My code prints (1 3 7 9) only first digits of numbers that should be printed. How to make it print all digits like in output?

CodePudding user response：

This works for every entered array, with one mistake that it always has one digit 1 surplus at the end.

#include <stdio.h>
#include <stdlib.h>
int subtract(int a[], int y, int k) {
  int count[100] = {0}, n[100], b[100][100], number[100], x;
  n[k] = a[k];
  while (n[k] != 0) {
    n[k] /= 10;
    count[k]  ;
  }
  number[k] = count[k];
  if (count[k] != 0) {
    count[k] = 0;
    n[k] = a[k];
    while (n[k] != 0) {
      b[k][count[k]] = n[k] % 10;
      n[k] /= 10;
      count[k]  ;
    }
  }
  int i;
  for (i = number[k] - 2; i >= 0; i--) {
    printf("%d", abs(b[k][i] - b[k][i   1]));
  }
}
int main() {
  int a[4] = {101, 3002, 70008, 900001}, n = 4, i;
  for (i = 0; i < n; i  ) {
    printf("%d ", subtract(a, n, i));
  }
  return 0;
}