C++ use class with conversion operator as index into array

Question

#include <cinttypes>
#include <type_traits>

template<typename Id, typename Value>
class sparse_set {

  static_assert(std::is_integral_v<Id>, "");  (1)
  static_assert(std::is_unsigned_v<Id>, "");

  Value& operator[](Id id);

  void push_back(const Value& value);
 
  // class implementation left out

};

class entity {

public:
  explicit entity(std::uint32_t id) : _id(id) {}
  ~entity() = default;

  std::uint32_t id() const {
    return _id;
  }

  operator std::uint32_t() const {  (2)
    return _id;
  }

private:
  std::uint32_t _id;

}; // class entity

int main() {
    const auto e = entity{2};

    auto set = sparse_set<entity, int>{};

    set.push_back(0);
    set.push_back(1);
    set.push_back(2);
    set.push_back(3);

    auto i = set[e];  (3)
    
    return 0;
}

I am trying to use a class with a conversion operator to std::uint32_t (2) as an index into a container class (3). Accessing an element with an instance of that class works and i get the right element.

But testing the class with a static_assert and std::is_unsigned_v and std::is_integral_v results in an assertion failure.

I need assertions to make sure Id can be used as an index.

When I static_assert with std::uint32_t everything works so I would expect the conversion operator to work aswell.

No, static_assert(std::is_integral_v<entity>); would complain. "Provides the member constant value which is equal to true, if T is the type bool, char, char8_t (since C++20), char16_t, char32_t, wchar_t, short, int, long, long long, or any implementation-defined extended integer types, including any signed, unsigned, and cv-qualified variants. Otherwise, value is equal to false.". Also note: "The behavior of a program that adds specializations for is_integral or is_integral_v (since C++17) is undefined." — Ted Lyngmo
– Ted Lyngmo, Commented Sep 21, 2021 at 10:58
your code has several errors not related to the question. Please fix them and also add the necessary includes — 463035818_is_not_an_ai
– 463035818_is_not_an_ai, Commented Sep 21, 2021 at 11:03

463035818_is_not_an_ai · Accepted Answer · 2021-09-21 11:05:11Z

4

entity certainly is not an integral type. It can be converted to a type that can be used as index and thats what you require:

#include <type_traits>

template<typename Id, typename Value>
class sparse_set {

  static_assert(std::is_convertible_v<Id,size_t>, "");

  // class implementation left out

};

answered Sep 21, 2021 at 11:05

463035818_is_not_an_ai

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Comments

Ruks · Accepted Answer · 2021-09-21 11:20:11Z

2

std::is_unsigned and std::is_integral only works for primitive types, it doesn't work for classes, even if they are implicitly convertible to a type they support. You can solve this by two ways:

Create your own trait:

#include <type_traits>

// ...

using uint16 = std::uint16_t;
using uint32 = std::uint32_t;

class entity { /* ... */ };

template <typename T>
struct is_valid_index : std::is_unsigned<T> {};
template <>
struct is_valid_index<entity> : std::true_type {};

template <typename T>
constexpr auto is_valid_index_v = is_valid_index<T>::value;

template<typename Id, typename Value>
class sparse_set {
    static_assert(is_valid_index_v<Id>, "");
    // ...
};

// ...

Remove std::is_unsigned completely and use std::is_convertible instead:

#include <type_traits>

// ...

using uint16 = std::uint16_t;
using uint32 = std::uint32_t;

class entity { /* ... */ };

template<typename Id, typename Value>
class sparse_set {
    static_assert(std::is_convertible_v<Id, uint32>, "");
    // ...
};

// ...

edited Sep 21, 2021 at 11:20

answered Sep 21, 2021 at 11:08

Ruks

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1 Comment

463035818_is_not_an_ai Over a year ago

isnt anything that is_unsigned also is_convertible_v<T,uint32> ?

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