Restrict variadic template arguments

Question

Can we restrict variadic template arguments to a certain type? I.e., achieve something like this (not real C++ of course):

struct X {};

auto foo(X... args)
         


        

Answer 1

Yes it is possible. First of all you need to decide if you want to accept only the type, or if you want to accept a implicitly convertible type. I use std::is_convertible in the examples because it better mimics the behavior of non-templated parameters, e.g. a long long parameter will accept an int argument. If for whatever reason you need just that type to be accepted, replace std::is_convertible with std:is_same (you might need to add std::remove_reference and std::remove_cv).

Unfortunately, in C++ narrowing conversion e.g. (long long to int and even double to int) are implicit conversions. And while in a classical setup you can get warnings when those occur, you don't get that with std::is_convertible. At least not at the call. You might get the warnings in the body of the function if you make such an assignment. But with a little trick we can get the error at the call site with templates too.

So without further ado here it goes:

---

The testing rig:

struct X {};
struct Derived : X {};
struct Y { operator X() { return {}; }};
struct Z {};

foo_x : function that accepts X arguments

int main ()
{
   int i{};
   X x{};
   Derived d{};
   Y y{};
   Z z{};

   foo_x(x, x, y, d); // should work
   foo_y(x, x, y, d, z); // should not work due to unrelated z
};

---

C++20 Concepts

Not here yet, but soon. Available in gcc trunk (March 2020). This is the most simple, clear, elegant and safe solution:

#include 

auto foo(std::convertible_to auto ... args) {}

foo(x, x, y, d); // OK
foo(x, x, y, d, z); // error:

We get a very nice error. Especially the

constraints not satisfied

is sweet:

Dealing with narrowing:

I didn't find a concept in the library so we need to create one:

template 
concept ConvertibleNoNarrowing = std::convertible_to
    && requires(void (*foo)(To), From f) {
        foo({f});
};

auto foo_ni(ConvertibleNoNarrowing auto ... args) {}

foo_ni(24, 12); // OK
foo_ni(24, (short)12); // OK
foo_ni(24, (long)12); // error
foo_ni(24, 12, 15.2); // error

C++17

We make use of the very nice fold expression:

template )>>
auto foo_x(Args... args) {}

foo_x(x, x, y, d, z);    // OK
foo_x(x, x, y, d, z, d); // error

Unfortunately we get a less clear error:

template argument deduction/substitution failed: [...]

Narrowing

We can avoid narrowing, but we have to cook a trait is_convertible_no_narrowing (maybe name it differently):

template 
struct is_convertible_no_narrowing_impl {
  template () = {std::declval()})>
  static auto test(F f, T t) -> std::true_type;
  static auto test(...) -> std::false_type;

  static constexpr bool value =
      decltype(test(std::declval(), std::declval()))::value;
};

template 
struct is_convertible_no_narrowing
    : std::integral_constant<
          bool, is_convertible_no_narrowing_impl::value> {};

C++14

We create a conjunction helper:
_{please note that in C++17 there will be a std::conjunction, but it will take std::integral_constant arguments}

template 
struct conjunction {};

template 
struct conjunction
    : std::integral_constant::value>{};

template 
struct conjunction : std::integral_constant {};

and now we can have our function:

template ::value...>::value>>
auto foo_x(Args... args) {}


foo_x(x, x, y, d); // OK
foo_x(x, x, y, d, z); // Error

C++11

just minor tweaks to the C++14 version:

template 
struct conjunction {};

template 
struct conjunction
    : std::integral_constant::value>{};

template 
struct conjunction : std::integral_constant {};

template ::value...>::value>::type>
auto foo_x(Args... args) -> void {}

foo_x(x, x, y, d); // OK
foo_x(x, x, y, d, z); // Error