在 class 模板的可变函数模板上使用 decltype 定义成员变量

Defining member variable using decltype on variadic function template of a class template

我正在尝试在可变函数模板上使用 'decltype' 来获取其 return 值类型,然后使用它来定义成员变量。但我不断收到此错误:

D:\Qt Projects\OpenGL_PhysicsSim\lib\Physics Effects\include\ParticleList.hpp:89: error: cannot convert 'std::tuple<std::uniform_real_distribution<double>, std::uniform_real_distribution<double>, std::uniform_real_distribution<double> >' to 'int' in assignment
     distributionTuple = createDistribution<Type, Types...>(meanValue, meanValues..., varianceValue, varianceValues...);

基本上,decltype 失败并将 distributionTuple 声明为 int 而不是推断出 createDistribution 的 return 类型。

template<typename AttributeType, typename Type, typename ...Types>
class ParticleAttributeGenerator
{
private:
 template<typename T>
 auto createDistribution(T meanValue, T varianceValue)
 {
   static_assert(
     std::is_integral<Type>::value || std::is_floating_point<Type>::value,
     "Type should be either integral value or floating point value");

   using distType = typename std::conditional< std::is_integral<T>::value,
                                               std::uniform_int_distribution<>,
                                               std::uniform_real_distribution<> >::type;
   T a = meanValue - varianceValue;
   T b = meanValue + varianceValue;

   return std::tuple<distType>(distType(a,b));
 }

 template<typename Tfirst, typename ...Trest>
 auto createDistribution(Tfirst meanValue, Trest... meanValues, Tfirst varianceValue, Trest... varianceValues)
 {
   static_assert(
     std::is_integral<Type>::value || std::is_floating_point<Type>::value,
     "Type should be either integral value or floating point value");

   using distType = typename std::conditional< std::is_integral<Tfirst>::value,
                                               std::uniform_int_distribution<>,
                                               std::uniform_real_distribution<> >::type;
   Tfirst a = meanValue - varianceValue;
   Tfirst b = meanValue + varianceValue;

   static_assert((sizeof...(meanValues)) == (sizeof...(varianceValues)), "number of meanValues and varianceValues should match!");

   return std::tuple_cat(std::tuple<distType>(distType(a,b)), createDistribution<Trest...>(meanValues..., varianceValues...));
 }

public:
 ParticleAttributeGenerator(Type meanValue, Types... meanValues, Type varianceValue, Types... varianceValues)
 {
   distributionTuple = createDistribution<Type, Types...>(meanValue, meanValues..., varianceValue, varianceValues...);  // 89 : error          
 }

private:
  //using result_type_t = typename std::result_of<createDistribution(Type, Types..., Type, Types...)>::type;
  decltype (createDistribution<Type, Types...>(Type, Types..., Type, Types...)) distributionTuple;
  //decltype (createDistribution<Type, Types...>(0.0f, 0.0f, 0.0f, 0.0f, 0.0f, 0.0f)) distributionTuple;
};

当我提供 createDistribution 的所有参数值时它起作用,但这不是我正在寻找的行为。因为我不知道该函数会有多少参数,所以它必须保留为可变参数模板函数。

我打算如何使用 class 模板的示例:

ParticleAttributeGenerator<glm::vec3, float, float, float> example1(3.0f, 1.0f, 3.0f, 1.0f, 3.0f, 1.0f);
ParticleAttributeGenerator<glm::u8vec4, uint8_t, uint8_t, uint8_t, uint8_t> example2(3, 2, 25, 5, 51, 12, 32, 3);

如果我没有任何成员变量并使用distributedTuple作为:

auto distributionTuple = createDistribution<Type, Types...>(meanValue, meanValues..., varianceValue, varianceValues...);

它编译,因此我相信 createDistribution 设法递归定义元组。但这对我没有用。一定有什么地方错了,我想念的。 我在 -std=c++14 模式下使用 GCC 4.9.2。

当您需要在未计算的上下文中构建函数的完整调用时,请使用 std::declval。使用它,您可以像这样声明 distributionTuple

decltype (std::declval<ParticleAttributeGenerator>().createDistribution<Type, Types...>(std::declval<Type>(), std::declval<Types>()..., std::declval<Type>(), std::declval<Types>()...)) distributionTuple;

首先,让我们将示例大幅简化为更易于管理且外部内容更少的内容:

template <typename T>
class Bar
{
private:
    template <typename U>
    auto foo(U a, U b)
    {
        return std::tuple<U>(a+b);
    }

public:
    Bar(T a, T b)
    {
        distributionTuple = foo<T>(a, b);
    }

private:
    decltype (foo<T>(T, T)) distributionTuple;
};

int main()
{
    Bar<int> b(4, 4);
}

这给了我你在问题中出现的示例编译错误(无法将 std::tuple<int> 转换为 int。那是因为:

foo<T>(T, T)

不是有效的函数调用。您需要使用具有这些类型的 表达式 调用 foo<T>。不仅仅是类型列表。为此,我们有 std::declval。即:

foo<T>(std::declval<T>(), std::declval<T>())

一旦你做了这个改变,你会得到一个新的编译错误:"cannot call member function foo without object." 所以让我们再次添加一个带有 declval 的对象。以下编译:

template <typename T>
class Bar
{
private:
    template <typename U>
    auto foo(U a, U b)
    {
        return std::tuple<U>(a+b);
    }

public:
    Bar(T a, T b)
    {
        distributionTuple = foo<T>(a, b);
    }

private:
    decltype(std::declval<Bar>().foo<T>(
                 std::declval<T>(), 
                 std::declval<T>())
             ) distributionTuple;
};