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类模板,在定义中引用它自己的类型

metaprogramming language-lawyer templates c++

rmawatson · 技术社区 · 6 年前

我想知道visual studio 2017和gcc中哪一个是正确的,关于标准,下面的情况是正确的。问题在于,在类模板second中,visual studio中的标识符second总是引用具体的类型,但在gcc中,它似乎依赖于上下文。

GCC示例

template <typename...>
struct type_list{};

template<template <typename...> typename tmpl> 
struct tmpl_c
{
    template <typename...Ts> using type = tmpl<Ts...>;
};

template<typename> struct template_of;
template <template <typename...> typename tmpl, typename... Ts>
struct template_of<tmpl<Ts...>>{
    using type = tmpl_c<tmpl>; 

};


template <typename T>
struct first{};


template <typename T>
struct second
{
    // 'second' here refers to second<int>
    using test = second; 

    // 'second' here refers to the template second
    // is this due to the context? ie that the tmpl_c is expecting a template not a concrete type?
    using types = type_list<tmpl_c<first>, tmpl_c<second> >; 


    // second here refers to the concrete type 'second<int>'
    // this workaround is needed for visual studio as it seems second always 
    // refers to the concrete type, not the template.
    using types2 = type_list<tmpl_c<first>, typename template_of<second>::type >; 
};

Demo

和visual studio示例(只有不同的位)

template <typename T>
struct second
{
    // 'second' here refers to second<int>
    using test = second; 

    // 'second' here refers to second<int>
    // this doesn't compile in visual studio as second<int> not the template.
    //using types = type_list<tmpl_c<first>, tmpl_c<second> >; 


    // second here refers to the concrete type 'second<int>'
    // this workaround is needed for visual studio as it seems second always 
    // refers to the concrete type, not the template.
    using types2 = type_list<tmpl_c<first>, typename template_of<second>::type >; 
};

Demo

作为 template_of 两种情况下的解决方案似乎都能正常工作,这是我目前唯一的选择。但我还是想知道哪个是正确的,或者是否还有其他的解决方法。

fixed in visual studio 15.8 Preview 2

1 回复 | 直到 6 年前

songyuanyao 6 年前

GCC是正确的。根据使用规则 injected-class-name for class template ,注入的类名也可以用作模板名或类型名。

(强调我的)

与其他类一样,类模板有一个注入的类名。注入的类名可以用作模板名或类型名。

在以下情况下,注入的类名被视为类模板本身的模板名:

后面跟着<

它用作对应于模板模板参数的模板参数

它是友元类模板声明的详细类说明符中的最终标识符。

否则,它将被视为类型名,并等效于模板名,后跟<>中包含的类模板的模板参数。

也就是说,

template <typename T>
struct second
{

    // second is same as second<T>
    using test = second; 

    // second is considered as a template-name
    // it's used as template argument for a template template parameter
    using types = type_list<tmpl_c<first>, tmpl_c<second> >; 

    // second is same as second<T>
    using types2 = type_list<tmpl_c<first>, typename template_of<second>::type >; 
};