Extract a number from a position in C

Question

In this exercise he asks me to create a function Number_pos (N, pos, m) which allows to extract a number composed of m digits from position pos using functions.

Example: N = 12345, pos = 2, m = 3
Number_pos (N, pos, m) = 234

I use an Extraxt_from_position function which extracts a number from a given position, then I use a function which calculates the number of digits of the number to extract, then I have a mirror function which inverts the number and I do the successive division until the number of digits are equal to the number of digits of the number we want to extract.

The problem is: forbidden to use mirror function, can you help me

int Extract_from_position(int n, int r)
{
  int m = 0, s = 0;
  while (n != 0)
  {
    m = n % 10;
    s++;
    if (s == r)
    {
      return n;
    }
    n = n / 10;
  }
}

int Number_of_digits(int n)
{
  int m = 0, s = 0;
  while (n != 0)
  {
    m = n % 10;
    s++;
    n = n / 10;
  }
  return s;
}

int Mirror(int n)
{
  int m = 0, s = 0;
  while (n != 0)
  {
    m = n % 10;
    s = s * 10 + m;
    n = n / 10;
  }
  return s;
}

int Number_Pos(int N, int pos, int m)
{
  int x = Extract_from_position(N, pos);
  int y = 0;
  int R = Mirror(x);
  int T = Number_of_digits(R);
  while (T >= m + 1)
  {
    y = R % 10;
    R = R / 10;
    T--;
  }
  return Mirror(R);
}

int main()
{
  int n, pos, nbcx;
  printf("Give n :");
  scanf("%d", &n);
  printf("Give the position :");
  scanf("%d", &pos);
  printf("give the number of digits of the number to extract :");
  scanf("%d", &nbcx);
  printf("\nThe result after the amber extract from position %d on the right and the number of digits %d is : %d  \n", pos, nbcx, Number_Pos(n, pos, nbcx));
}

Answer 1

UPDATE: Count from right
If you want count the digits from right, the NumberPos function will be just:

#include <stdio.h>
#include <math.h>

int NumberPos(int N, int pos, int m)
{
    int trc = (int)(N / (int)pow(10, pos - 1));
    trc = trc % (int)pow(10, m);
    return trc;
}

int main()
{
  int n, pos, nbcx;
  printf("Give n :");
  scanf("%d", &n);
  printf("Give the position :");
  scanf("%d", &pos);
  printf("give the number of digits of the number to extract :");
  scanf("%d", &nbcx);
  printf("\nThe result after the amber extract from position %d on the right and the number of digits %d is : %d  \n", pos, nbcx, NumberPos(n, pos, nbcx));
}

And the output will be, for example:

Give n :1234567 
Give the position :3
give the number of digits of the number to extract :2

The result after the amber extract from position 3 on the right and the number of digits 2 is : 45 

---

OLD: This could be a solution (in basically 4 line):

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

int NumberPos(int N, int pos, int m)
{
    int digit = floor(log10(abs(N))) + 1;
    int trc = N % (int)pow(10, digit - pos + 1);
    digit = floor(log10(abs(trc))) + 1;
    trc = (int)(trc / (int)pow(10, digit - m));
    return trc;
}

int main()
{
  int n, pos, nbcx;
  printf("Give n :");
  scanf("%d", &n);
  printf("Give the position :");
  scanf("%d", &pos);
  printf("give the number of digits of the number to extract :");
  scanf("%d", &nbcx);
  printf("\nThe result after the amber extract from position %d on the right and the number of digits %d is : %d  \n", pos, nbcx, NumberPos(n, pos, nbcx));
}

The output will be:

Give n :12345
Give the position :2
give the number of digits of the number to extract :2

The result after the amber extract from position 2 on the right and the number of digits 2 is : 23 

---

UPDATE: Library restriction
If for whatever reason you are not allowed to use math.h or stdlib.h you can:

• Re-implement pow reading: Write Pow Function Without math.h in C
• Re-implement abs reading: this
• Re-implement the digit counter: C program to count number of digits in an integer

Answer 2

Something like this might work. I trim the digits you don't want on the right then mod to mask off the digits on the left you don't want.

Based on your sample I assume that pos is 1-based. If not there's a comment on the code you would need to remove.

You'd probably want to add error checking to make sure that pos and num_digits are valid for the given N, but that's an exercise for you.

#include <stdio.h>

int Number_of_digits(int n)
{
    int count = 0;
    while (n != 0)
    {
        n = n / 10;
        ++count;
    }
    return count;
}

int Number_Pos(int N, int pos, int num_digits)
{
    int len = Number_of_digits(N);
    pos -= 1; //pos is 1 based.

    //trim right side
    for (int i = 0; i < len - num_digits - pos; ++i)
    {
        N /= 10;
    }

    //calculate mod to keep num_digits.
    int m = 10;
    for (int i = 0; i < num_digits - 1; ++i)
    {
        m *= 10;
    }

    return N % m;
}

int main()
{
    int n = 1234567;
    int pos = 2;
    int num_digits = 3;
    int result = Number_Pos(n, pos, num_digits);
    printf("Num: %d, Pos: %d, Digits: %d - Result: %d\n", n, pos, num_digits, result);

    pos = 3;
    num_digits = 4;
    result = Number_Pos(n, pos, num_digits);
    printf("Num: %d, Pos: %d, Digits: %d - Result: %d\n", n, pos, num_digits, result);

    pos = 1;
    num_digits = 4;
    result = Number_Pos(n, pos, num_digits);
    printf("Num: %d, Pos: %d, Digits: %d - Result: %d\n", n, pos, num_digits, result);

    pos = 6;
    num_digits = 2;
    result = Number_Pos(n, pos, num_digits);
    printf("Num: %d, Pos: %d, Digits: %d - Result: %d\n", n, pos, num_digits, result);

    return 0;
}

Output:

Num: 1234567, Pos: 2, Digits: 3 - Result: 234
Num: 1234567, Pos: 3, Digits: 4 - Result: 3456
Num: 1234567, Pos: 1, Digits: 4 - Result: 1234
Num: 1234567, Pos: 6, Digits: 2 - Result: 67