Task

Write a program that solves the odd word problem with the restrictions given below.



Description

You are promised an input stream consisting of English letters and punctuations.

It is guaranteed that:

 the words (sequence of consecutive letters) are delimited by one and only one punctuation,
 the stream will begin with a word,
 the words will be at least one letter long,   and 
 a full stop (a period, [.]) appears after, and only after, the last word.


Example

A stream with six words:

 what,is,the;meaning,of:life. 


The task is to reverse the letters in every other word while leaving punctuations intact, producing:

 what,si,the;gninaem,of:efil. 
while observing the following restrictions:

 Only I/O allowed is reading or writing one character at a time, which means: no reading in a string, no peeking ahead, no pushing characters back into the stream, and no storing characters in a global variable for later use;
 You are not to explicitly save characters in a collection data structure, such as arrays, strings, hash tables, etc, for later reversal;
 You are allowed to use recursions, closures, continuations, threads, co-routines, etc., even if their use implies the storage of multiple characters.


Test cases

Work on both the   "life"   example given above, and also the text:

 we,are;not,in,kansas;any,more.


#include <iostream>
#include <cctype>
#include <functional>
 
using namespace std;
 
bool odd()
{
  function<void ()> prev = []{};
  while(true) {
    int c = cin.get();
    if (!isalpha(c)) {
      prev();
      cout.put(c);
      return c != '.';
    }
    prev = [=] { cout.put(c); prev();  };
  }
}
 
bool even() 
{
  while(true) {
    int c;
    cout.put(c = cin.get());
    if (!isalpha(c)) return c != '.';
  }
}
 
 
int main()
{
  bool e = false;
  while( e ? odd() : even() ) e = !e;
  return 0;
}