TITLE: word counting example using STL

(Newsgroups: comp.lang.c++, 24 May 2000)

STRANGWICK: (craig_strangwick@hotmail.com)

: 1. words are completeley made of alphabetic, numeric, hyphens
: and apostrophe's.

KUEHL: dietmar.kuehl@claas-solutions.de

OK, this is easy to accomplish.

STRANGWICK:

: 2. apostrophe's appearing before AND after a word should be
: ignored.

KUEHL:

That is, individual apostrophe's like in "a ' a" should not count as a
word but embedded apostrophe's like in "don't" should not separate
words, is this correct? Probably the same applies to hyphens. Here is
the trivial code:

  #include <locale>
  #include <algorithm>
  #include <iostream>
  #include <string>

  class space_ctype: public std::ctype<char>
  {
  public:
    space_ctype(): std::ctype<char>(get_table()) {}

  private:
    static void set_space(std::ctype_base::mask& c) { c |= space; }
    static std::ctype_base::mask* get_table() {
      static std::ctype_base::mask rc[table_size];
      std::copy(classic_table(), classic_table() + table_size, rc);
      std::for_each(rc, rc + table_size, set_space);
      static char letters[] = "abcdefghijklmnopqrstuvwxyz"
                              "ABCDEFGHIJKLMNOPQRSTUVWXYZ"
                              "0123456789-'";
      for (char* tmp = letters; tmp != letters + sizeof(letters); )
        rc[*tmp++] &= ~space;
      return rc;
    }
  };

  int main() {
    std::locale l;
    std::locale new_l(l, new space_ctype);
    std::cin.imbue(new_l);
    std::string str;

    int count = 0;
    while (std::cin >> str)
      if (str.find_first_not_of("-'") != str.npos)
        ++count;
    std::cout << "words: " << count << "\n";
  }

If there wasn't this restriction with respect to hyphens and apostrophe's,
the code would be somewhat simpler: The "if" could be dropped :-)


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