Thread: Could someone explain this program?

It's in my book, but unexplained. It reverses a 2-digit number. But I don't get how the operations work....Shouldn't it give zero each time? Could someone please explain what's happening?

Code:

#include<stdio.h>

int main(void)
{
   int number, last_digit, next_digit, rev_number;

  printf("Enter the number to be reversed:");
  scanf("%d", &number);

  last_digit=number-((number/10)*10);

  rev_number=last_digit;

  next_digit=(number/10) - ((number/100)*10);

  rev_number=(rev_number*10)+next_digit;

  printf("The reversed number is:%d", rev_number);
  getch();
  clrscr();

}

Did you run it?
What was the result?

I also have a feeling, that, this is incorrect, and way too complex also.

Assume number =532

lastdigit=532-((532/10)*10=532-53*10=2

Shouldn't it give zero each time?

Note that integer/integer is always rounded off to integer (in programming). So 532/10=53.2=53

Note that integer/integer is always rounded off to integer (in programming). So 532/10=53.2=53

Oh, all right, then I get it... Thanks!