Re: Practicing recursion

From:

Michael DOUBEZ <michael.doubez@free.fr>

Newsgroups:

comp.lang.c++

Date:

Tue, 14 Apr 2009 14:37:44 +0200

Message-ID:

<49e4810a$0$15223$426a34cc@news.free.fr>

Fencer wrote:

Hello, need to do some exercices that involves recursion, the
substitution model in particular (I believe that what it's called). I
decided to translate the following lisp program that sums the elements
in a list that are numbers:

(defun is-integer (x) ; Helper function to sum-only-numbers
  (typep x 'integer)
)

(defun sum-only-numbers (list)
   (cond
    ((endp list) 0)
    ((is-integer (first list)) (+ (first list) (sum-only-numbers(rest
list))))
    (t (sum-only-numbers (rest list)))
    )
)

(sum-only-numbers '(1 2 a b 33)) ; which evals to 36

I don't really know lisp and I'm interested in C++ right now, just
showing it for references.

Here's my attempt:
#include <iostream>
#include <list>
#include <sstream>
#include <string>

using namespace std;

bool is_num(const string& s)
{
   istringstream iss(s);

   int n;

   return iss >> n;
}

int to_num(const string& s)
{
   istringstream iss(s);

   int n;

   iss >> n;

   return n;
}

The is_num and to_num could be merged into the faillible class idiom:

template<class T>
struct faillible
{
   T value;
   bool success;
   faillible(const T& v, bool s=false) : value(v), success(s) { }
};

faillible<int> to_num(const string& s)
{
  faillible<int> result(0);
  //...
  result.success=(iss>>result.value);
  return result;
}

int sum_only_numbers(list<string>::const_iterator itr,
list<string>::const_iterator end)
{
if (itr != end)
{

    const faillible<int> num_conversion=to_num(*itr);
    if(num_conversion.success)
    {
       return num_conversion.value + sum_only_numbers(++itr, end);
    }
    else
    {
       return sum_only_numbers(++itr, end);
    }

   else
   {
      return 0;
   }
}

int main(void)
{
   string ss[] = {"1", "2", "a", "b", "33"};
   list<string> l(ss, ss + sizeof(ss) / sizeof(ss[0]));

   cout << sum_only_numbers(l.begin(), l.end()) << endl;
}

It prints "36" when run so it seems to work! Any helpful suggestions
before I hand it in?

--
Michael