Overload assignment operator for assigning sql::ResultSet to struct tm

Question

Are there exceptions for types which can't have thier assignment operator overloaded?

Specifically, I'm wanting to overload the assignment operator of a struct tm (from time.h) so I can assign a sql::ResultSet to it.

I already have the conversion logic:

sscanf(sqlresult->getString("StoredAt").c_str(), "%d-%d-%d %d:%d:%d",
  &TempTimeStruct->tm_year, &TempTimeStruct->tm_mon, &TempTimeStruct->tm_mday,
  &TempTimeStruct->tm_hour, &TempTimeStruct->tm_min, &TempTimeStruct->tm_sec);

I tried the overload with this:

tm& tm::operator=(sql::ResultSet & results)
{
  /*CODE*/
  return *this;
}

However VS08 reports:

error C2511: 'tm &tm::operator =(sql::ResultSet &)' : overloaded member function not found in 'tm'

Answer 1

The assignment operator must be a member function (of struct tm in this case), so the only way of doing this would be to modify the standard library itself, something you should definitely not do. You can of course write a named free function to do whatever you want.

Answer 2

struct mytm : public tm {
  mytm& operator =(tm const& input) {
     /* whatever you want it to do. */
  return *this;
};

Some folks don't think much of this. I'll attach the following caveat:

If you got a ton of tm items (or any other classic c-lib struct) floating around in your code, and you want to take control over them, wrapping a class around them this way allows you do so. Just doing it for a 3-line knockoff is probably not a great idea. It's possible that the critics would prefer:

struct mytm {
  tm the_tm;
  /* etc */
};

Instead.

Answer 3

tm to_tm(sql::ResultSet const& results) {
  tm ret;
  int r = sscanf(results->getString("StoredAt").c_str(), "%d-%d-%d %d:%d:%d",
    &ret->tm_year, &ret->tm_mon, &ret->tm_mday,
    &ret->tm_hour, &ret->tm_min, &ret->tm_sec);
  if (r != 6) throw std::runtime_error("sscanf: unexpected results");
  return ret;
}

void example() {
  tm t = to_tm(results);
}