c++ - 根据模板类型选择类模板的成员类型？

Question

我有一个包含某种类型成员的类模板。该类型是根据实例化模板时提供的类型确定的。它使用默认值(在下面的示例中为 double)，除非该类提供了覆盖。用作模板类型的类可以提供此覆盖 type (此处“Two”提供覆盖类型“int”)。如果某个类提供了覆盖，则仅当该类还设置了 UseOverride 标志时才应使用覆盖。如果标志不存在或为假，则应使用默认的“double”。

问题是，如果模板类型未提供“类型”，则编译器会在下面的代码中给出错误。我怀疑我需要在这里使用 SFINAE，但即使在下午的大部分时间里困惑和浏览相关问题之后，仍无法找到合适的方法。

如何定义 EventType 模板以使其按预期工作？我想保留EventType<T>语法。

#include <iostream>

struct One {
    //This type is ignored, and double is used, because UseOverride = true is not specified:
    using type = short;
};
struct Two {
    static constexpr bool UseOverride = true;
    using type = int;
};

struct Three {
    static constexpr bool UseOverride = false;
    //I don't want the compiler to complain that "type" is not available here (because it should default to double anyhow since 
    //the class instructs not to use the override). But compile does generate error. 
    //How to avoid this?
};

template <typename T, typename = void>
struct overrideInfoProvided : std::false_type {};
template <typename T>
struct overrideInfoProvided<T, decltype((void)T::UseOverride, void())> : std::true_type {};

template <typename T>
constexpr bool Override()
{
    if constexpr (overrideInfoProvided<T>::value)
    {
        return T::UseOverride;
    }
    return false;
}

template<class T>
using EventType = typename std::conditional_t<Override<T>(), typename T::type, double>;


template <class T>
struct Test
{
    typename EventType<T> member;
    Test()
    {
        std::cout << member << std::endl;
    }
};

int main()
{
    Test<One>();
    Test<Two>();   
    //Gives error:
    //Test<Three>();// `type': is not a member of any direct or indirect base class of `three'; 
}

Answer 1

I don't want the compiler to complain that "type" is not available here (because it should default to double anyhow since the class instructs not to use the override). But compiler does generate error. How to avoid this?

只需使用以下 type_identity 推迟对 ::type 的访问即可助手:

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

template <typename T>
using EventType = typename std::conditional_t<Override<T>()
                                            , T
                                            , type_identity<double>>::type;
//                                            ~~~~~~~~~~~~^        

DEMO