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/* I want to overload (+,-,%,*,/) to the class mult*/ ...

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Question by kimo3333
Submitted on 6/14/2004
Related FAQ: comp.lang.c Answers (Abridged) to Frequently Asked Questions (FAQ)
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/* I want to overload (+,-,%,*,/) to the class mult*/
#include<iostream.h>
#include<conio.h>
#include<string.h>

#define TRUE 1
#define FALSE 0

#define NEXT TRUE
#define LAST FALSE

char *chstr(char ,char*);

class digit;
typedef unsigned char byte;
typedef digit *dgtp;

class digit
{
friend class mult;
friend class multscan;
private:
     static int radix;//to be used by the all classes
     byte dgt;
     dgtp next;
     dgtp last;

public:
     digit (int);
     friend ostream  &operator<<(ostream &s, const digit &d);//overloading of iostream to be used with the class/
     friend istream  &operator>>(istream &s,  digit &d);
};
int digit ::radix=256;

digit::digit(int i)
{
        dgt =(byte)i ;
        next=last=0 ;
}

  ostream &operator<<(ostream &s, const digit &d)
        {
        s << int(d.dgt);
        return s;
        }
        istream &operator>>(istream &s, digit &d)
        {   int i;
           s >> i;
          if (i<0 ||i >= digit::radix)
            //"enter a byte in op>>(os&,dig&)");
            d.dgt = (byte)i ;
            return s;
        };
    const int POSV=+1;
    const int NEGV=-1;
  class mult
          {
          friend class multscan;
          private:
                    int sgn;
                    dgtp head;
                    dgtp tail;
                    int numdgt;
              void  clearmult();
         public:
                    mult ()   {sgn=POSV;head= tail=0;numdgt=0;}
                    mult (int);
                    mult(mult&);
                    ~mult()   {clearmult(); }  ;
         void setsgn (char);
         char getsign ();
         void chsgn();
         void prepend(int);
         void append(int);
         int gethead();
         int gettail();
         int lastdec();
         mult& subdec(int);
         mult& divten ();
         char *decimalize();
         mult &operator= ( mult &) ;

};
void mult::clearmult()
{
  if (!head)return;
  dgtp dp =head;
  while (dp)
    {
      dgtp ddp =dp;
      if ((dp=dp->next) !=0)dp->last= 0;
      delete ddp;
    }   ;
  head=tail=0;
  numdgt=0;
}
mult::mult(int i)
{
  if (i<0||i>=digit::radix) {
  //  error ("enter a byte!");
  sgn=POSV;
  head=tail=new digit(i);
  numdgt=1;};
}
void mult::setsgn(char ch)
{
  if (ch == '+') sgn = POSV;
  else if (ch == '-') sgn = NEGV;
//else cerr("enter '+' or '-'only!");
}
char mult::getsign(){return (sgn==POSV) ?'+' :'-';}
void mult::chsgn() {sgn =(sgn ==POSV) ?NEGV :POSV;}
/*the isertion function are for head insertion*/
void mult::prepend (int i)
{
  dgtp p=new digit(i);
if (!p)   // error("allocation error in mult::prepend (int)");
if(!head)head=tail=p;else
{ head->last=p;
  p->next=head;
  head=p;
} ;
  ++numdgt;
}
/*and for tail isertion*/
void mult::append(int i)
{
  dgtp p=new digit(i);
  //if (!p)
  //error("allocation error in mult::append (int)");
  if (!tail)
     head =tail=p;
  else
    {
      tail->next=p;
      p->last=tail;
      tail=p;
    }
    ++numdgt;

}

int mult::gethead ()
{
//if (!head)
  // error ("can't get from empty list in mult::gethead()");
  dgtp p=head;
  int i=int(p->dgt);
  if ((head=head->next)!=0)
  head->last=0;
  delete p;
  --numdgt;
  return i;
};
int mult::gettail()
{
if (!tail)
   cout<<"can't get from empty list in mult::gettail()";
  dgtp p=tail;
  int i=int(p->dgt);
  if ((tail=tail->last)!=0)tail->next=0;

  delete p;
  --numdgt;
  return i;
}
/*the copy constructor may be written in terms of prepend or append;we choose the latter*/
mult::mult (mult &m)
{
  head=tail=0;
  sgn=m.sgn;
  numdgt=0;
  while (m.head) append(m.gethead());
}
/*next we need input and output functions;these needn't be member functions.the input function is one which assemblesa list of digits*/
void makemult(mult &m)
  {
  char ch;
  cout<<"enter sign: ";
  cin>>ch;
  m.setsgn(ch);
  int n;
  cout<<"number of bytes? : ";
  cin>>n;
  for (int i=0;i<n;++i)
    {
     int b;
     cout<<"enter byte: ";
     cin>>b;
     m.append(b);
    };
  } ;

  class multscan
{
private:
        mult *mp;
        dgtp dp;
        bool dir;
public:
        multscan(mult&,bool);
        dgtp operator()();
};
/*ere sep is defined in standard.h to be the string constant", "and '\b'representsthebackspace character.
the heart of the problem of representing large integers as lists of bytes ,however,is the conversion of the multiple
byte integer so constructed into a normal integer>>>>>>>>>>>>>>*/
int mult::lastdec()
{
  int d;
  int B = digit::radix;
  int c = B%10;
  multscan thisptr(*this,NEXT);
  d=0;
  dgtp p;
  while ((p = thisptr())!= 0)
       d =(c*d + (p->dgt%10))%10;
  return d;
}
mult& mult::subdec(int i)
{
  int B = digit::radix;
  bool carry=FALSE;
  multscan thisptr (*this,LAST);
  dgtp p=thisptr();
  if (!p)
     cout<<"cann'tsubtract from empty list"<<"\n";
  int j=int(p->dgt)-i;
  if (j<0)
    {
     j +=B;
     carry=TRUE;
    };
  p->dgt=(byte)j;
  while((p = thisptr()) !=0)
    {
     j=int (p->dgt);
     if (carry)
       {
        --j;
        carry =FALSE;
       };
     if (j<0)
       {
        j +=B;
        carry =TRUE;
       };
     p->dgt =(byte)j;
    }
    if (!head->dgt) gethead();
    return *this;
}
/*we may now divide by 10.knowing that there will be no overall remainder,carrying the stepwise remainder,and deleting the head digit if its data value is zero*/
  mult& mult::divten()
   {
    int B=digit::radix;
    int c=0;
    multscan thisptr(*this,NEXT);
    dgtp p;
    while ((p =thisptr()) !=0)
      {
       int i = int (p->dgt);
       i +=(c*B);
       p->dgt =(byte)(i/10);
       c =i %10;
      }  ;
      if (!head->dgt) gethead();
      return *this;
   }

/*printing out the string which results yields decimal digits in the normal order with the aid of the dewcimal output member function (which also procedes the number which its sign):*/
char* mult::decimalize()
{
  char *number = "";
  while (head)
       {
           int d = lastdec();
           char ch = char(d+ '0');
           number = chstr(ch,number);
           subdec(d);
           if (head) divten() ;
          } ;
    char s = (sgn == POSV) ? '+' : '-';
    number = chstr(s,number);
    return number;

}

multscan::multscan (mult &m ,bool direction)
{
       mp = &m;
       dir=direction;
       dp=dir ? m.head : m.tail;
}
dgtp multscan::operator()()
{
  dgtp p=dp;
  if (dir) dp = p ? dp->next:0;
  else  dp = p ? dp->last:0;
  return p;
}

/*and the ouput function should be one which prints out a multiple byte integer in some suitable format*/
void printmult (mult &m)
{
  multscan thisptr(m,NEXT);
  cout<<"("<<m.getsign();
  dgtp p;
  while ((p=thisptr()) !=0)
     cout<<*p<<", ";
  cout<<"\b\b)";
  } ;
/*here sep is defined in standard.h to be the string constant", "and '\b'representsthebackspace character.
the heart of the problem of representing large integers as lists of bytes ,however,is the conversion of the multiple
byte integer so constructed into a normal integer>>>>>>>>>>>>>>*/
/*if we now repeat the process we obtain the second to last decimal digit:and so on .
However,if we print out the decimals as they arrive,they will be in inverse order.It is therfore necessary to ouput the character value of each decimal digit in turnk to the head of a character string. For a C string:*/
char *chstr(char ch,char *str)
   {
    int n = strlen(str);
    char *newstr = new char[n+2];
    newstr[0] = ch;
    newstr[1]='\0';
    str = strcat(newstr,str);
    //newstr = 0;
    return str;
   } ;

mult &mult::operator=(mult &m)
{
clearmult();
while (m.tail)append(m.gethead()) ;

return *this;
}
void main()
{
  mult m;
  makemult(m);
  printmult(m);
  cout<<"\n";
  cout<<m.decimalize()<<"\n";
  getch();
}


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