Flaming Dangerzone

Beating overload resolution into submission

a guest post by Xeo

Battle Intro

When dealing with overloaded templates, it's often desirable to restrict certain overloads so that they don't accidentally get selected. A previous blog post already explains how to do that, the disadvantage being that you need to have non-overlapping conditions (or disjoint conditions). Let's take a silly example:

template<unsigned N, EnableIf<is_multiple_of<N, 3>>...>
void print_fizzbuzz(){ std::cout << "fizz\n"; }

template<unsigned N, EnableIf<is_multiple_of<N, 5>>...>
void print_fizzbuzz(){ std::cout << "buzz\n"; }

template<unsigned N, EnableIf<is_multiple_of<N, 15>>...>
void print_fizzbuzz(){ std::cout << "fizzbuzz\n"; }

template<unsigned N, DisableIf<Any<is_multiple_of<N, 3>, is_multiple_of<N, 5>>>...>
void print_fizzbuzz(){ std::cout << N << "\n"; }

A call to print_fizzbuzz<15>() will be ambiguous, since the first three overloads are all viable and enabled — after all, 15 is a multiple of 3, 5 and 15 itself. So, how do we go about disambiguating these?

Domestic Abuse

Why, by abusing a language feature of course! And one that is in quite a close relationship with overloading, to boot: Implicit conversions. I could bore you with a slew of standardese at this point, but I think I'll just summarize: overloads with the least amount of implicit conversions on each argument, if any, are preferred.

The fix is pretty well known — overloading on int and long, and passing a literal 0 as the argument. This is often used when defining a trait's test functions:

template<class T>
struct some_trait{
private:
  template<class U /*, possibly some SFINAE test here */>
  static auto test(int)
      -> decltype(/* possibly some Expression SFINAE test here */, void());
  template<class>
  static char test(long);
  
public:
  static constexpr value = std::is_void<decltype(test<T>(0))>::value;
};

(There is an in-depth example / explanation on Stack Overflow.)

The literal 0 is of type int, as such the call test<T>(0) will prefer the first overload, if it's viable (e.g. not disabled via SFINAE). If that one is not viable, the second overload will provide a fallback because int is implicitly convertible to long. Perfect.

Your Princess Is In Another Castle

... or not, since it only allows to disambiguate two overloads. It also can't be used in a binary fashion, as overload resolution will look at each argument in isolation to determine ambiguity. It is, however, a good starting point — we can impose a total order over an unordered overload set with this. Let's see how we can increase the number of overloads.

There's one type of parameter that, by definition, will be selected as the very last choice for implicit conversions — the ellipsis (C-style variadic functions). With this, we can now order a total of three overloads, enough for our original problem:

template<unsigned N, EnableIf<is_multiple_of<N, 15>>...>
void print_fizzbuzz(int){ std::cout << "fizzbuzz\n"; }

template<unsigned N, EnableIf<is_multiple_of<N, 3>>...>
void print_fizzbuzz(long){ std::cout << "fizz\n"; }

template<unsigned N, EnableIf<is_multiple_of<N, 5>>...>
void print_fizzbuzz(...){ std::cout << "buzz\n"; }

template<unsigned N, DisableIf<Any<is_multiple_of<N, 3>, is_multiple_of<N, 5>>>...>
void print_fizzbuzz(...){ std::cout << N << "\n"; }

We also add the ellipsis to the fourth overload, since it needs to take something, otherwise it would be ruled out thanks to not having enough parameters. So the original call would now be print_fizzbuzz<15>(0), working all fine and dandy. However, I think seeing the types directly in the overload is more noise than anything, so we should fix that with a few using-aliases and a type for the ellipsis:

using choice_1 = int;
using choice_2 = long;

struct otherwise{ otherwise(...){} };

template<unsigned N, EnableIf<is_multiple_of<N, 15>>...>
void print_fizzbuzz(choice_1){ std::cout << "fizzbuzz\n"; }

template<unsigned N, EnableIf<is_multiple_of<N, 3>>...>
void print_fizzbuzz(choice_2){ std::cout << "fizz\n"; }

template<unsigned N, EnableIf<is_multiple_of<N, 5>>...>
void print_fizzbuzz(otherwise){ std::cout << "buzz\n"; }

template<unsigned N, DisableIf<Any<is_multiple_of<N, 3>, is_multiple_of<N, 5>>>...>
void print_fizzbuzz(otherwise){ std::cout << N << "\n"; }

otherwise just wraps the ellipsis behind another conversion, which allows us to give it a nice name.

For Great Justice

There's something that still bugs me, though — the need for DisableIf to make the fourth overload, which is otherwise unrestricted, disjoint from the first three. And if we can impose a total order with three overloads, we can surely get it to four, right?

Turns out we can go even farther than that. To do that, we have to take a step back and not only look at implicit conversions of unqualified types (like int and long), but also at how qualifications play into this. C++ actually provides us with a total ordering of references to cv-qualified types already, with every "extra" cv-qualification making the overload less preferred. Last but not least, we have any kind of rvalue reference being preferred to an lvalue reference-to-const for rvalue arguments.

using choice_1 = int&&;
using choice_2 = int const&&;
using choice_3 = int const volatile&&;
using choice_4 = int const&;

Yielding the following overload set.

void f(choice_1){}
void f(choice_2){}
void f(choice_3){}
void f(choice_4){}
void f(otherwise);

The above overloads would all be unambiguous when called as f(0), no matter how many of them exist. That alone doesn't get us much farther than before, but if we add long into the mix...

using choice_5 = long&&;
using choice_6 = long const&&;
using choice_7 = long const volatile&&;
using choice_8 = long const&;

... and adding the respective overloads to f, we can suddenly order up to nine overloads! Overload resolution see's the rvalue int argument and tries to find a fitting overload. If there are no immediately matching overloads for int (e.g., they where all not viable), it tries the same implicit conversion as before, producing an rvalue long. That again is tried to be matched against a fitting overload. If there isn't a fitting one this time, we go for the least preferred overload, the C-style variadic one. Nice.

This in effect means that we can now remove the DisableIf restriction on the general print_fizzbuzz case, leaving us with this:

// I put this one on top to show the order we want them invoked in
template<unsigned N, EnableIf<is_multiple_of<N, 15>>...>
void print_fizzbuzz(choice_1){ std::cout << "fizzbuzz\n"; }

template<unsigned N, EnableIf<is_multiple_of<N, 3>>...>
void print_fizzbuzz(choice_2){ std::cout << "fizz\n"; }

template<unsigned N, EnableIf<is_multiple_of<N, 5>>...>
void print_fizzbuzz(choice_3){ std::cout << "buzz\n"; }

template<unsigned N>
void print_fizzbuzz(otherwise){ std::cout << N << "\n"; }

Which looks nice and clean, defining clearly how we want the restrictions to apply, and without needing to basically invert all restrictions for the general case. (Yes, I know, the long overloads weren't necessary for this case, but they came for free anyways.)

Up To Eleven Thirteen

So, the obvious question is: can we extend this even more? Yes, abusing not only standard conversion sequences (SCS), like we have until now, but also involving user-defined conversion sequences (UCS). An SCS consists of things like integer promotion, cv-qualification conversion, array-to-pointer decay, basically things that work without you needing to do anything. A UCS consists of an SCS, followed by an implicit user-defined conversion (constructor, conversion operator), followed by another SCS. We're going to abuse the second SCS here.

struct ranker{ operator int(){ return 0; } };

Yep, that's it. Just add another four overloads, netting us a total of thirteen.

// Note that all these are *more* preferred than the `int` and `long` ones
// so the numbers on those should be bumped
using choice_1 = ranker&&;
using choice_2 = ranker const&&;
using choice_3 = ranker const volatile&&;
using choice_4 = ranker const&;

With the call now being f<15>(ranker{}). Overload resolution now first tries to match the ranker argument, then tries its implicit (user-defined) conversion to int, then the implicit (standard) conversion from that int to long, and at the end, if everything fails, chooses the ellipsis. Awesome.

A Riddle Wrapped In A Mystery Inside An Enigma

Time to tackle the first SCS. The first SCS is basically used to determine which conversion operation is going to be performed (i.e., which constructor or conversion operator is going to be called). If you have a T& object for example, with an operator U() const; and are converting to U, the first SCS will consist of a qualification adjustment to T const&, after which the conversion operator will be called.

The first idea that springs to mind might be the following, remembering the non-const/const ranking we had before:

struct choice_2{};
using choice_3 = int;
using choice_4 = long;

struct choice_1{
  operator choice_2(){ return {}; }
  operator choice_3() const{ return 0; }
};

// for looks
struct select_overload : choice_1{};

With a smaller example:

void f(choice_1){}
void f(choice_2){}
void f(choice_3){}
void f(choice_4){}

And the call being f(select_overload{}). Unfortunately, this won't work, because of the standard being a sissy some arcane rules saying that the first SCS doesn't count for the ordering, unless it's a derived-to-base conversion. Well, if they say so, we can arrange for that:

using choice_3 = int;
using choice_4 = long;

struct choice_2{
  operator choice_3(){ return 0; }
};
struct choice_1 : choice_2{};

With the same overloads and the same invokation, this actually works. If a choice_1 can't be matched, it will converted to choice_2 and thrown into the matching-pit again. If that fails too, the base class' operator choice_3() is considered, which requires a derived-to-base conversion as the first SCS, fulfilling the ordering criteria from before. After that, it's normal ranking as we had before. If we throw qualifications back into the mix, we get a grand total of seventeen overloads.

But Wait, There's More!

Astute readers will have noticed that the final ranking code contains the Secret Ingredient to an even more general solution. It just so turns out that C++ has another total ordering of conversion built-in — an infinitely extensible one to boot. I'm talking about the derived-to-base conversion of course, which we considered a kind of a hinderence last time.

template<unsigned I>
struct choice : choice<I+1>{};

// terminate recursive inheritence at a convenient point,
// large enough to cover all cases
template<> struct choice<10>{};

// I like it for clarity
struct select_overload : choice<0>{};

Yep, that's it. Looks deceivingly simple, but is very powerful. Let's adjust our fizzbuzz and for the kick, add a few extra cases:

template<unsigned N, EnableIf<is_multiple_of<N, 15>>...>
void print_fizzbuzz(choice<0>){ std::cout << "fizzbuzz\n"; }

template<unsigned N, EnableIf<is_multiple_of<N, 21>>...>
void print_fizzbuzz(choice<1>){ std::cout << "fizzbeep\n"; }

template<unsigned N, EnableIf<is_multiple_of<N, 27>>...>
void print_fizzbuzz(choice<2>){ std::cout << "fizznarf\n"; }

template<unsigned N, EnableIf<is_multiple_of<N, 35>>...>
void print_fizzbuzz(choice<3>){ std::cout << "buzzbeep\n"; }

template<unsigned N, EnableIf<is_multiple_of<N, 45>>...>
void print_fizzbuzz(choice<4>){ std::cout << "buzznarf\n"; }

template<unsigned N, EnableIf<is_multiple_of<N, 63>>...>
void print_fizzbuzz(choice<5>){ std::cout << "beepnarf\n"; }

template<unsigned N, EnableIf<is_multiple_of<N, 3>>...>
void print_fizzbuzz(choice<6>){ std::cout << "fizz\n"; }

template<unsigned N, EnableIf<is_multiple_of<N, 5>>...>
void print_fizzbuzz(choice<7>){ std::cout << "buzz\n"; }

template<unsigned N, EnableIf<is_multiple_of<N, 7>>...>
void print_fizzbuzz(choice<8>){ std::cout << "beep\n"; }

template<unsigned N, EnableIf<is_multiple_of<N, 9>>...>
void print_fizzbuzz(choice<9>){ std::cout << "narf\n"; }

// we stay with `otherwise` for the general case, so we
// don't have to adjust anything if new overloads are added
template<unsigned N>
void print_fizzbuzz(otherwise){ std::cout << N << "\n"; }

With the magic call being the obvious: print_fizzbuzz<15>(select_overload{});. And it works, too. Overload resolution is now an obedient little dog with a cute collar on. Mission accomplished, I'd say.


Special thanks go to Johannes Schaub - litb for helping me figure out the last part. Thanks to Luc Danton for otherwise.