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)

Here my intention is to have a function which accepts a variable number of X parameters.

The closest we have is this:

template <class... Args>
auto foo(Args... args)

but this accepts any type of parameter.

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 <concepts>

auto foo(std::convertible_to<X> 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 <class From, class To>
concept ConvertibleNoNarrowing = std::convertible_to<From, To>
    && requires(void (*foo)(To), From f) {
        foo({f});
};

auto foo_ni(ConvertibleNoNarrowing<int> 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 <class... Args,
         class Enable = std::enable_if_t<(... && std::is_convertible_v<Args, X>)>>
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 <class From, class To>
struct is_convertible_no_narrowing_impl {
  template <class F, class T,
            class Enable = decltype(std::declval<T &>() = {std::declval<F>()})>
  static auto test(F f, T t) -> std::true_type;
  static auto test(...) -> std::false_type;

  static constexpr bool value =
      decltype(test(std::declval<From>(), std::declval<To>()))::value;
};

template <class From, class To>
struct is_convertible_no_narrowing
    : std::integral_constant<
          bool, is_convertible_no_narrowing_impl<From, To>::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 <bool... B>
struct conjunction {};

template <bool Head, bool... Tail>
struct conjunction<Head, Tail...>
    : std::integral_constant<bool, Head && conjunction<Tail...>::value>{};

template <bool B>
struct conjunction<B> : std::integral_constant<bool, B> {};

and now we can have our function:

template <class... Args,
          class Enable = std::enable_if_t<
              conjunction<std::is_convertible<Args, X>::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 <bool... B>
struct conjunction {};

template <bool Head, bool... Tail>
struct conjunction<Head, Tail...>
    : std::integral_constant<bool, Head && conjunction<Tail...>::value>{};

template <bool B>
struct conjunction<B> : std::integral_constant<bool, B> {};

template <class... Args,
          class Enable = typename std::enable_if<
              conjunction<std::is_convertible<Args, X>::value...>::value>::type>
auto foo_x(Args... args) -> void {}

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

Answer 2

What about the following solution ?

--- EDIT --- Improved following suggestion from bolov and Jarod42 (thanks!)

#include <iostream>

template <typename ... Args>
auto foo(Args... args) = delete;

auto foo ()
 { return 0; }

template <typename ... Args>
auto foo (int i, Args ... args)
 { return i + foo(args...); }

int main () 
 {
   std::cout << foo(1, 2, 3, 4) << std::endl;  // compile because all args are int
   //std::cout << foo(1, 2L, 3, 4) << std::endl; // error because 2L is long

   return 0;
 }

You can declare foo() to receive all types of arguments (Args ... args) but (recursively) implement it only for one type (int in this example).

Answer 3

How about static_assert and helper template method (c++11 solution):

template <bool b>
int assert_impl() {
   static_assert(b, "not convertable");
   return 0;
}

template <class... Args>
void foo_x(Args... args) {
    int arr[] {assert_impl<std::is_convertible<Args, X>::value>()...};
    (void)arr;
}

One more c++11 this one uses "one-liner" sfinae-based solution:

template <class... Args,
          class Enable = decltype(std::array<int, sizeof...(Args)>{typename std::enable_if<std::is_convertible<Args, X>::value, int>::type{}...})>
void foo_x(Args... args) {
}

Answer 4

C++14

Since C++14 you can use also variable template, partial specialization and static_assert to do that. As an example:

#include <type_traits>

template<template<typename...> class, typename...>
constexpr bool check = true;

template<template<typename...> class C, typename U, typename T, typename... O>
constexpr bool check<C, U, T, O...> = C<T, U>::value && check<C, U, O...>;

template<typename... T>
void f() {
    // use std::is_convertible or whichever is the best trait for your check
    static_assert(check<std::is_convertible, int, T...>, "!");
    // ...
}

struct S {};

int main() {
    f<int, unsigned int, int>();
    // this won't work, for S is not convertible to int
    // f<int, S, int>();
}

You can also use check in conjunction with std::enable_if_t as a return type, if you don't want to use static_assert for some unknown reasons:

template<typename... T>
std::enable_if_t<check<std::is_convertible, int, T...>>
f() {
    // ...
}

And so on...

C++11

In C++11, you can also design a solution that stops the recursion immediately when a type that is not to be accepted is encountered. As an example:

#include <type_traits>

template<bool...> struct check;
template<bool... b> struct check<false, b...>: std::false_type {};
template<bool... b> struct check<true, b...>: check<b...> {};
template<> struct check<>: std::true_type {};

template<typename... T>
void f() {
    // use std::is_convertible or whichever is the best trait for your check
    static_assert(check<std::is_convertible<int, T>::value...>::value, "!");
    // ...
}

struct S {};

int main() {
    f<int, unsigned int, int>();
    // this won't work, for S is not convertible to int
    // f<int, S, int>();
}

As mentioned above, you can use check also in the return type or wherever you want.

Answer 5

You already have it since C++11 standard.

A simple std::array (special case of std::tuple where all the tuple elements share the same type) will be sufficient.

However, if you want to use it in a template function, you may better use an ´std::initializer_list` like in the following example:

template< typename T >
void foo( std::initializer_list<T> elements );

This is a really simple solution which solves your problem. Using variadic template arguments is an option too, but adds unnecessary complexity to your code. Remember that your code should be readable by others, including yourself after some time.

Answer 6

The one aspect of this question that @463035818_is_not_a_number's answer does not address is how to avoid creating gratuitous copies of arguments. Here's an idea, due to this blog post, that minimizes copying arguments that are expensive to copy:

struct X { void use() { /* ... */ } /* ... */ };

template<typename Want, typename Have>
inline std::conditional_t<std::is_same_v<Want, Have>, Want &&, Want>
local_copy(Have &in)
{
  return static_cast<Have&&>(in);
}

template<std::convertible_to<X> ...T>
void
foo1(T&&...t)
{
  // Unary fold over comma operator
  (local_copy<X, T>(t).use(), ...);
}

// Another way to do it
template<std::convertible_to<X> ...T>
void
foo2(T&&...t)
{
  auto use = []<typename U>(U &&arg) {
    decltype(auto) x = local_copy<X, U>(arg);
    x.use();
  };
  (use(std::forward<T>(t)), ...);
}

The key idea is that local_copy will return a copy of its argument unless the argument was an rvalue of exactly the desired type, in which case it will return an rvalue reference to its argument. Hence, in the event that you call, say:

X x;
foo1(X{}, x);

The first argument, X{} will construct a temporary object before foo1's invocation, and foo1 can just modify this temporary in place and never copy it. By contrast, the second argument x gets copied, ensuring any modifications to the argument stay local to the function foo1.