Extract a number from a position in C - 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