在可变参数模板中初始化静态数组 class

Question

我有一个可变参数模板 class "InterleavedAttribute"，现在我想要一些关于模板中类型的运行时信息 ("AttributeTypeInfo")。我想在模板 class ("attributeInfo") 中填充一个静态 const 数组，然后在运行时使用索引访问该数组，例如 object.typeInfo(2)，但我无法弄清楚如何。这是我拥有的：

#include <iostream>

template <typename TYPE> struct GLTypeInfo;

template <>
struct GLTypeInfo<char> { static constexpr unsigned int glType = 0; };

template <>
struct GLTypeInfo<int> { static constexpr unsigned int glType = 1; };

template <>
struct GLTypeInfo<float> { static constexpr unsigned int glType = 2; };

//---------------------------------------------------------------------------

struct AttributeTypeInfo
{
    unsigned int nrOfComponents;
    unsigned int glType;
    unsigned int bytesPerComponent;
    unsigned int bytesPerAttribute;

    constexpr AttributeTypeInfo(unsigned int _nrOfComponents, unsigned int _glType, unsigned int _bytesPerComponent, unsigned int _bytesPerAttribute)
        : nrOfComponents(_nrOfComponents), glType(_glType), bytesPerComponent(_bytesPerComponent), bytesPerAttribute(_bytesPerAttribute)
    {}       
};

template <typename T>
struct TypeInfoFactory
{
    static unsigned int constexpr extent = std::is_array<T>::value ? std::extent<T>::value : 1;
    using type = typename std::conditional<std::is_array<T>::value, typename std::remove_extent<T>::type, T>::type;
    static constexpr AttributeTypeInfo typeInfo = AttributeTypeInfo(extent, GLTypeInfo<type>::glType, sizeof(type), sizeof(type) * extent);
};

//---------------------------------------------------------------------------

template <typename TYPE, typename ...P> struct InterleavedAttribute
{
    static constexpr unsigned int nrOfAttributes = sizeof...(P)+1;
    //static constexpr AttributeTypeInfo attributeInfo[sizeof...(P)+1] = { ??? }; // <-- How do I fill this...
    TYPE data;
    InterleavedAttribute<P...> next;
    InterleavedAttribute() {}
    InterleavedAttribute(const TYPE & value, const P &... p) : data(value), next(p...) {}
    //static constexpr AttributeTypeInfo typeInfo(unsigned int index) { return attributeInfo[index]; } // ...so I can call this later?
};

template <typename TYPE> struct InterleavedAttribute<TYPE>
{
    static constexpr unsigned int nrOfAttributes = 1;
    static constexpr AttributeTypeInfo attributeInfo[1] = { TypeInfoFactory<TYPE>::typeInfo };
    TYPE data;
    InterleavedAttribute() {}
    InterleavedAttribute(const TYPE & value) : data(value) {}
    static constexpr AttributeTypeInfo typeInfo(unsigned int index) { return attributeInfo[0]; }
};

int main() {
    InterleavedAttribute<int> ia1(5);
    std::cout << "Numer of attributes: " << ia1.nrOfAttributes << std::endl;
    std::cout << "Attribute 0, glType: " << ia1.typeInfo(0).glType << std::endl;

    InterleavedAttribute<float, int, char> ia3(1.2, 3, 'a');
    std::cout << "Numer of attributes: " << ia3.nrOfAttributes << std::endl;
    //std::cout << "Attribute 0, glType: " << ia3.typeInfo(0).glType << std::endl; <-- Trying to get type information here
}

代码在 Coliru here。我找到了 this 答案，而且很接近，但我仍然无法真正弄清楚如何得到我想要的...

Answer 1

您可以使用 std::tuple 执行以下操作：

template <typename ...Ts> struct InterleavedAttribute
{
    static constexpr unsigned int nrOfAttributes = sizeof...(Ts);
    static constexpr AttributeTypeInfo attributeInfo[sizeof...(Ts)] = {
        TypeInfoFactory<Ts>::typeInfo...
    };
    std::tuple<Ts...> data;

    InterleavedAttribute() {}
    InterleavedAttribute(const Ts&... args) : data(args...) {}
    static constexpr AttributeTypeInfo typeInfo(unsigned int index) {
        return attributeInfo[index];
    }

    template <unsigned int I>
    auto attribute() -> decltype(std::get<I>(data)) { return std::get<I>(data); }
    template <unsigned int I>
    auto attribute() const -> decltype(std::get<I>(data)) { return std::get<I>(data); }
};

// odr-used, so definition required for linker.
template <typename ...Ts> 
constexpr AttributeTypeInfo InterleavedAttribute<Ts...>::attributeInfo[sizeof...(Ts)];

Demo

在可变参数模板中初始化静态数组 class

Initialize static array in variadic template class

c++

variadic-templates

c++11