Restoring Division Algorithm Implementation In C

This is a dynamic program for the implementation of the Restoring division Algorithm in the C. Restoring Division uses Left Shift Operations, Twos’ Compliment and Binary Addition.

#include<stdio.h>

#include<malloc.h>

int * a, * q, * m, * mc, * c, n, d;
int powr(int x, int y) {
  int s = 1, i;
  for (i = 0; i < y; i++)
    s = s * x;
  return s;
}
void print(int arr[], int n) {
  int i;
  for (i = 0; i < n; i++)
    printf("%d ", arr[i]);
}
void bin(int n, int arr[]) {
  int r, i = 0;
  do {
    r = n % 2;
    n /= 2;
    arr[i] = r;
    i++;
  } while (n > 0);
}
void set(int array[], int x) {
  int i, tmp[20] = {
    0
  };
  for (i = x - 1; i >= 0; i--)
    tmp[x - 1 - i] = array[i];
  for (i = 0; i < x; i++)
    array[i] = tmp[i];
}
int len(int x) {
  int i = 0;
  while (powr(2, i) <= x) i++;
  return ++i;
}

void addBinary(int a1[], int a2[]) {
  int bi[2] = {
    0
  }, ca[20] = {
    0
  };
  int t = len(n), tmp = 0;
  int * su = (int * ) malloc(sizeof(int) * len(n));
  while (t-- > 0) {
    tmp = a1[t] + a2[t] + ca[t];
    bin(tmp, bi);
    su[t] = bi[0];
    ca[t - 1] = bi[1];
    bi[0] = 0;
    bi[1] = 0;
  }
  for (t = 0; t < len(n); t++)
    a1[t] = su[t];
  free(su);
}

void twoCom(int arr[]) {
  int i;
  int * one = (int * ) malloc(sizeof(int) * len(n));
  for (i = 0; i < len(n) - 1; i++)
    one[i] = 0;
  one[i] = 1;
  for (i = 0; i < len(n); i++) {
    arr[i] = 1 - arr[i];
  }
  addBinary(arr, one);
  free(one);
}
void ls(int alen, int blen) {
  int i = 0;
  for (i = 0; i < alen - 1; i++)
    a[i] = a[i + 1];
  a[i] = q[0];
  for (i = 0; i < blen - 1; i++)
    q[i] = q[i + 1];
  q[i] = -1;
}
void printaq() {
  print(a, len(n));
  printf("t");
  print(q, len(n) - 1);
  printf("t");
  printf("n");
}
int main() {
  int i, cnt = 0;
  printf("Enter The Numerator/Denominator: ");
  scanf("%d/%d", & n, & d);
  q = (int * ) malloc(sizeof(int) * len(n) - 1);
  bin(n, q);
  m = (int * ) malloc(sizeof(int) * (len(n)));
  bin(d, m);
  a = (int * ) malloc(sizeof(int) * (len(n)));
  for (i = 0; i < len(n); i++)
    a[i] = 0;
  mc = (int * ) malloc(sizeof(int) * (len(n)));
  bin(d, mc);
  set(q, len(n) - 1);
  set(m, len(n));
  set(mc, len(n));
  twoCom(mc);
  cnt = len(n) - 1;
  printf("t Att Qtt Mt Countn");
  printf("t---------t-------tt--------- ---------n");
  while (cnt > 0) {
    printf("t");
    print(a, len(n));
    printf("t");
    print(q, len(n) - 1);
    printf("t");
    print(m, len(n));
    printf("t%dn", cnt);
    ls(len(n), len(n) - 1);
    printf("LSHIFTt");
    printaq();
    addBinary(a, mc);
    printf("A=A-Mt");
    printaq();
    if (a[0] == 1) {
      q[len(n) - 2] = 0;
      addBinary(a, m);
    } else
      q[len(n) - 2] = 1;
    printf("A=A+Mt");
    printaq();
    cnt -= 1;
  }
  return 0;
}

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