如何在构造函数中应用算术类型提升

Question

假设，如果我正在创建一个点 class 并且我希望它根据参数推导出类型，我希望它将点 class 提升到最高参数。例如：

template <class dtype>
class Point;

...

auto x = Point(1, 1.0); // Point<double> specialized
auto y = Point(1.0, 1); // Point<double> specialized

我不确定如何在构造函数中实现这一点。我已经能够让它从一个调用显式专用构造函数的函数中推断出类型，而不是从构造函数本身推断出类型。

这是我目前的尝试：

#include <type_traits>

template <typename... Ts>
struct promoted_type_wrap;

template <typename T>
struct promoted_type_wrap<T> {
  using type = T;
};

template <typename T, typename U, typename... Ts>
struct promoted_type_wrap<T, U, Ts...> {
  using type = typename promoted_type_wrap<typename std::conditional<
    (sizeof(U) <= sizeof(T)), T, U >::type, Ts... >::type;
};

template <typename... Ts>
using promoted_type = typename promoted_type_wrap<Ts...>::type;

template <typename T>
using same_type = typename promoted_type_wrap<T>::type;

template <class dtype>
class Point {
protected:
  dtype x, y;

public:
  constexpr Point(const dtype x, const same_type<dtype> y)
    : x(x), y(y) {
  }
};

template <class dtype, class etype>
constexpr auto make_Point(const dtype x, const etype y) {
  return Point<promoted_type<dtype, etype>>(x, y);
}

void test() {
  constexpr auto x = make_Point(1, 2.0); // Point<double> specialized
  constexpr auto y = make_Point(1.0, 2); // Point<double> specialized
  constexpr auto z = Point(1, 2.0); // Point<int> specialized
  constexpr auto w = Point(1.0, 2); // Point<double> specialized
}

为什么 Point(1, 2.0) 被专门化为 Point<int> 是有道理的，因为第一个参数是 int，它强制构造函数中的第二个参数是 int；但是，我不确定如何重写构造函数以使其表现得像伪构造函数工厂。

Answer 1

however, I am unsure how to go about rewriting the constructor to behave like the pseudo-constructor factory.

不是构造函数：您必须编写自定义推导指南。

内容如下

template <typename T1, typename T2>
Point(T1, T2) -> Point<promoted_type<T1, T2>>;

下面是一个完整的编译示例

#include <type_traits>

template <typename... Ts>
struct promoted_type_wrap;

template <typename T>
struct promoted_type_wrap<T>
 { using type = T; };

template <typename T, typename U, typename... Ts>
struct promoted_type_wrap<T, U, Ts...>
 { using type = typename promoted_type_wrap<std::conditional_t<
      (sizeof(U) <= sizeof(T)), T, U >, Ts... >::type; };

template <typename... Ts>
using promoted_type = typename promoted_type_wrap<Ts...>::type;

template <typename dtype>
class Point
 {
   protected:
      dtype x, y;

   public:
      template <typename T1, typename T2>
      constexpr Point (T1 const & a, T2 const & b) : x(a), y(b)
       { }
 };

template <typename T1, typename T2>
Point(T1, T2) -> Point<promoted_type<T1, T2>>;

int main ()
 {
   constexpr auto z = Point(1, 2.0); // now Point<double>
   constexpr auto w = Point(1.0, 2); // again Point<double> 

   static_assert( std::is_same_v<decltype(z), Point<double> const> );
   static_assert( std::is_same_v<decltype(w), Point<double> const> );
 }

题外话：我认为这不是一个好主意select "promoted type" 根据你[=19 中的类型大小=]

template <typename T, typename U, typename... Ts>
struct promoted_type_wrap<T, U, Ts...>
 { using type = typename promoted_type_wrap<std::conditional_t<
      (sizeof(U) <= sizeof(T)), T, U >, Ts... >::type; };

即使忽略其他问题，当您有相同大小的不同类型时，selected 类型是第一个。

例如，在我的平台中，g++ 和 clang++ 都有 sizeof(long) == sizeof(float)，所以我们得到

constexpr auto z = Point(1l, 2.0); // <-- deduced as Point<long>
constexpr auto w = Point(1.0, 2l); // <-- deduced as Point<double>

static_assert( std::is_same_v<decltype(z), Point<long> const> );
static_assert( std::is_same_v<decltype(w), Point<double> const> );

我建议使用 select "preferred type" 独立于类型顺序的东西。

在我看来你应该使用 std::common_type 如下

#include <type_traits>

template <typename dtype>
class Point
 {
   protected:
      dtype x, y;

   public:
      template <typename T1, typename T2>
      constexpr Point (T1 const & a, T2 const & b) : x(a), y(b)
       { }
 };

template <typename T1, typename T2>
Point(T1, T2) -> Point<std::common_type_t<T1, T2>>;

int main ()
 {
   constexpr auto z = Point(1l, 2.0); // <-- deduced as Point<double>
   constexpr auto w = Point(1.0, 2l); // <-- deduced as Point<double>

   static_assert( std::is_same_v<decltype(z), Point<double> const> );
   static_assert( std::is_same_v<decltype(w), Point<double> const> );
 }

如何在构造函数中应用算术类型提升

How to apply arithmetic type promotion within a constructor

templates

constructor

template-argument-deduction

type-promotion

c++17