Code snippet: easy lexicographical ordering

The C++ standard library provides the std::lexicographical_compare function which takes two iterator ranges and returns a boolean, indicating which of the ranges is smaller. But what if we have a data structure like this:

struct foo
  int a;
  std::string b;
  char c;

  bool operator<(foo const &) const;

for which we want to write an operator< which does a lexicographical comparison of a, b and c. We end up with the following:

bool foo::operator<(foo const &r) const
  if (a < r.a)
    return true;
  if (r.a < a)
    return false;
  if (b < r.b)
    return true;
  if (r.b < b)
    return false;
  return c < r.c;

In case you’re wondering, I’m assuming we work on types which only have an operator<, which is why I cannot use operator> and operator!=.

Now, this is tedious to write and error prone. And it can be solved more elegantly. You might know that std::pair has an operator< which does lexicographical comparison. So I checked if boost::tuple has this feature, too. As it turns out, it does! So the above code can be reduced to:

bool foo::operator<(foo const &r) const
  return boost::make_tuple(a,b,c) < boost::make_tuple(r.a,r.b,r.c);

Note that you need the tuple/tuple_comparison.hpp header for this to work. I checked the latest C++0x draft and it confirms that this feature is ported to C++0x, too. And if you’re lucky, all of the tuple stuff in the above code will be optimized away, so the handwritten code from above is left.

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