Rewrite the following functions and predicates in C using re

Rewrite the following functions and predicates in C++, using recursion if possible.

Functions and predicates on lists. my_first([H|_]) = H. my_last([X]) = X. my_last([_|T]) = my_last(T). contains([E|_], E) => true. contains([_|T], E) => contains(T, E). find_first_of(L, E) = find_first_of(L, E, 1). find_first_of([E|_], E, I) = I. find_first_of([_|T], E, I) = find_first_of(T, E, I+1). find_first_of([], _E, _I) = -1. find_last_of(L, E) = find_last_of(L, E, 1, -1). find_last_of([]PreI) = PreI. find_last_of([E|T], E, I, _PreI) = find_last_of(T, E, I+1, I). find_last_of([_|T], E, I, PreI) = find_last_of(T, E, I+1, PreI). % the kth_elm(L, K) is the same as L[K] kth_elm([E|_], 1) = E. kth_elm([_|T], K) = kth_elm(T, K-1). my_len([]) = 0. my_len([_|T]) = my_len(T)+1. my_reverse([]) = []. my_reverse([H|T]) = my_reverse(T) ++ [H], main => test_arith, test_list. test_arith => X = succ(succ(succ(0))), % 3 Y = succ (succ(0))y % 2 printf(\"%w + %w = %w\ \", X, Y, add(X, Y)), printf(\"%w - %w = %w\ \", X, Y, sub(X, Y)), printf(\"%w * %w = %w\ \", X, Y, mul(X, Y)), printf(\"%w/%w = %w\ \", X, Y, my_div(X, Y)), printf(\"%w %% %vi = %w\ \", X, Y, my_rem(X, Y)), printf(\"%w^%w = %w\ \", X, Y, exp(X, Y)), printf(\"&w;! = %w\ \", X, fact(X)), printf(\"gcd(%w, %w) = &w;\ \", X, Y, my_gcd(X, Y)), printf(\"fib(%w) = %w\ \", X, fib(X)). test_list => L = [a, b, c, a, b], writef(\"first(%w) = %w\ \", L, my_first(L)), writef(\"last(%w) = %w\ \", L, my_last(L)), if contains(L, a) then writef(\"%w contains %w.\ \", L, a) end, writef(\"find_first_of(%w, %w) = %w\ \", L, b, find_first_of(L, b)), writef(\"find_last_of(%w, %w) = %w\ \", L, b, find_last_of(L, b)), writef(\"kth(%w, %w) = %w\ \", L, 3, kth_elm(L, 3)), writef(\"len(%w) = %w\ \", L, my_len(L)), writef(\"reverse(%w) = %w\ \", L, my_reverse(L)).

Solution

#include<stdio.h>   
#include<iostream.h>
#include<conio.h>
class rtnl
{
   int nmr;
   int dnmm;
   public:
   void getdata()
   {
       cout<<\"\ enter the numerator part of the rtnl no.\";
       cin>>nmr;
       cout<<\"\ enter the denominator part of the rtnl no.\";
       cin>>dnmm;
   }
   void oprtr+(rtnl);
   void oprtr-(rtnl);
   void oprtr *(rtnl);
   void oprtr /(rtnl);
};
void rtnl ::oprtr+(rtnl c1)
{
   rtnl tmp;
   tmp.nmr=(nmr*c1.dnmm)+(c1.nmr*dnmm);
   tmp.dnmm=dnmm*c1.dnmm;
   cout<<\"\ rational no. after addition\";
   cout<<\"\ numerator=\"<<tmp.nmr<<\"\ denominator =\"<<tmp.dnmm;
}
void rtnl ::oprtr -(rtnl c1)
{
   rtnl tmp;
   tmp.nmr=(nmr*c1.dnmm)-(c1.nmr*dnmm);
   tmp.dnmm=dnmm*c1.dnmm;
   cout<<\"\ rtnl no. after subtraction\";
   cout<<\"\ numerator=\"<<tmp.nmr<,\"\ denominator =\"<<tmp.dnmm;
}
void rtnl ::oprtr (rtnl c1)
{
   rtnl tmp;
   tmp.nmr=nmr*c1.nmr;
   tmp.dnmm=dnmm*c1.dnmm;
   cout<<\"\ rtnl no. after multiplication\";
   cout <<\"\ numerator=\"<tmp.nmr<<\"\ denominator =\"<< tmp.dnmm;
}
void rtnl :: oprtr /(rtnl c1)
{
   rtnl tmp;
   tmp.nmr= nmr*c1.dnmm;
   tmp.dnmm=c1.nmr*dnmm;
   cout<<\"\ rtnl no. after dividation\";
   cout <<\"\ numerator=\"<<tmp.nmr<<\"\ denominator =\"<<tmp.dnmm;
}
void main()
{
   clrscr();
   rtnl c1, c2;
   int n;
   do
   {
       cout<<\"\ 1.Input data for rtnl no. \";
       cout<<\"\ 2. Addition of rtnl no. \";
       cout<<\"\ 3. Subtraction of rtnl no. \";
       cout<<\"\ 4. Multiplication of rtnl no.\";
       cout<<\ 5. Division of rtnl no. \";
       cout<<\"\ 6. Quit\";
       cout<<\"\ Enter your choice\";
       cin>>n;
       switch(n)
       {
           case 1:
           cout<<endl<<\"\ enter the data for first rtnl no.\";
           c1.getdata();
           cout<<endl<<\"\ enter the data for second rtnl no. \";
           c2.getdata ();
           clrscr();
           break;
           case 2;
           c1+c2;
           getch();
           clrscr();
           break;
           case 3;
           c1-c2;
           getch();
           clrscr();
           case 4:
           c1*c2;
           getch();
           clrscr();
           break;
           case 5:
           c1/c2;
           getch();
           clrscr();
           break;
           case 6:
           exit(1);
           break;
       }
   } while (n!=6);
   getch();
}