method that return digit by index

Question

I want to write a method that returns digit of a number (num) by index (place) while not using array only loops and conditions.

For example:

For the input num = 17489, place = 1 the method returns 9
But if the index does not exist in the like num = 17489, place = 6 return -1

something in my code doesn't work if I set i = 0 case 1) work but 2) no if i set i = 1 case 2) work but 1)no please help figure whats wrong

    public static int Digit_by_place(int num, int place)
    {
        int digit = 0, i = 0;

        // the lop run until the place in the 
        while (num > 0 && i <= place)number
        {
            digit = num % 10;
            num = num / 10;
            i++;

            // if there is no didgit in the input place return -1
            if (i < place && num == 0)
            {
                return -1;
            }

            // return the last digit from left (the bigest)
            if (num == 0 && digit > 0 && i == place)
            {
                return digit;
            }
        }

        return digit;
    }

A basic understanding of english language would be really helpful. It´s really hard to understand what works and what not. Apart from this please clearify what "not working" actually means. — MakePeaceGreatAgain, Dec 21 '18 at 11:14
@HimBromBeere in my code if i set i=1 and input any index between 1-num length it works the problem is if i input iindex>num length for example 14879,5 return 1 cause thats the last digit in the num but if i input 14879,6 it still return 1 insted of -1 — MarcoPolo, Dec 21 '18 at 11:38

Dmitry Bychenko · Answer 1 · 2018-12-21T11:59:25.740

We can exploit the fact that int can have at most 10 digits.

Code:

public static int Digit_by_place(int num, int place) {
  if (place <= 0 || place > 10)   // place must be in [1..10] range
    return -1;
  else if (num == 0)              // special case num == 0
    return place == 1 ? 0 : -1;

  for (int i = 1; i < place; ++i)  
    num /= 10;

  return num == 0 ? -1 : Math.Abs(num % 10);
}

Demo:

using System.Linq;
...

Tuple<int, int>[] tests = new Tuple<int, int>[] {
  Tuple.Create( 17489,  1),
  Tuple.Create( 17489,  2),
  Tuple.Create( 17489,  3),
  Tuple.Create( 17489,  4),
  Tuple.Create( 17489,  5),
  Tuple.Create( 17489,  6),
  Tuple.Create(-17489,  1),
  Tuple.Create(-17489,  6),
  Tuple.Create( 17489,  0),
  Tuple.Create( 17489, -1),
  Tuple.Create(-17489, -1),
  Tuple.Create( 17489,  5),
  Tuple.Create(-17489,  5),
  Tuple.Create(     0,  1),
  Tuple.Create(     0,  4),
};

var report = string.Join(Environment.NewLine, tests
  .Select(test => 
     $"Digit_by_place({test.Item1,6}, {test.Item2,2}) = {Digit_by_place(test.Item1, test.Item2),2}"));

Console.Write(report);

Outcome:

Digit_by_place( 17489,  1) =  9
Digit_by_place( 17489,  2) =  8
Digit_by_place( 17489,  3) =  4
Digit_by_place( 17489,  4) =  7
Digit_by_place( 17489,  5) =  1
Digit_by_place( 17489,  6) = -1
Digit_by_place(-17489,  1) =  9
Digit_by_place(-17489,  6) = -1
Digit_by_place( 17489,  0) = -1
Digit_by_place( 17489, -1) = -1
Digit_by_place(-17489, -1) = -1
Digit_by_place( 17489,  5) =  1
Digit_by_place(-17489,  5) =  1
Digit_by_place(     0,  1) =  0
Digit_by_place(     0,  4) = -1

@Magnus: I assume digits be in `[0..9]` otherwise `-1` which now stands for *error* will conflict with `-1` from, say `Digit_by_place(-17489, 5)`. If *signed* digits are required, we should remove `Math.Abs` — Dmitry Bychenko, Dec 21 '18 at 12:30
Yeh, I guess that makes sense if -1 is used for out of range — Magnus, Dec 21 '18 at 12:35

Magnus · Answer 2 · 2018-12-21T12:57:11.423

1

Using mathematics only.

public int GetNumber(int number, int position)
{
    if(number == 0 && position == 1) 
       return 0;
    if(position <= 0 || position > (int)(Math.Log10(number = Math.Abs(number)) + 1)) 
       return -1;
    return (int)((number % (Math.Pow(10, position))) / (Math.Pow(10, position - 1)));
}

edited Dec 21 '18 at 12:57

answered Dec 21 '18 at 12:21

Magnus

45,362
8
80
118

Now it's correct, +1; however using `double` math methods (`Math.Log10` and `Math.Pow`) when solving pure `integer` task is IMHO an overshoot. – Dmitry Bychenko Dec 21 '18 at 13:15

GrabnaMax · Answer 3 · 2018-12-21T11:51:56.847

0

for(int i=0; i<place; i++)
{
       num = num / 10;
       if (num <= 0)
           return -1;
}
return num % 10;

this should work if you start to count with 0

edited Dec 21 '18 at 11:51

answered Dec 21 '18 at 11:21

GrabnaMax

47
7

that's not work good it return 8 if i input 17489,1 insted of return 9 – MarcoPolo Dec 21 '18 at 11:32
if `num` is *non negative* integer (like `123`) condition `if (num < 0)` will be *never* met – Dmitry Bychenko Dec 21 '18 at 11:43

method that return digit by index

3 Answers3