c++ - std::ranges 算法函数对象自定义点 AKA niebloids - std lib 类型的问题

Answer 1

You need to specify a comparison for T if std::less<T> is unavailable.

Both std::sort and std::ranges::sort require a strict weak ordering predicate, not operator<=> (but that can supply one, via < and std::less)

template<typename T>
bool complex_less (std::complex<T> lhs, std::complex<T> rhs) { return abs(lhs) < abs(rhs); };

int main() {
  auto things = std::vector<thing>{{9, 10}, {7, 8}, {3, 4}, {1, 2}, {5, 6}};
  std::sort(things.begin(), things.end());
  std::ranges::sort(things);
  std::ranges::sort(things, {}, [](const auto& e) { return e.sum(); });
  std::ranges::sort(things, [](const auto& a, const auto& b) { return a < b; }, {});
  std::ranges::sort(things, {}, &thing::x);
  std::ranges::sort(things, {}, static_cast<int (thing::*)()>(&thing::sum)); // need cast to disambiguate pointer-to-member
  for (const auto& e: things) std::cout << e << " ";
  std::cout << "\n";

  auto complexes = std::vector<std::complex<double>>{{9, 10}, {7, 8}, {3, 4}, {1, 2}, {5, 6}};
  std::sort(complexes.begin(), complexes.end(), complex_less<double>); // fine
  std::ranges::sort(complexes, complex_less<double>);                  // also fine
  std::ranges::sort(complexes, {}, [](const auto& c) { return std::abs(c); }); // still fine
  std::ranges::sort(complexes, {}, [](const auto& c) { return c.real(); }); // fine
  for (const auto& e: complexes) std::cout << e << " ";
  std::cout << "\n";

  return EXIT_SUCCESS;
}