c++ - typedef 为模板模板的别名

Question

我有一个特征类 AttributeTraits，它使用智能指针模板类 MySmartPointer:

template<typename T>
class MySmartPointer {};

template<template <typename> class SP = MySmartPointer>
struct AttributeTraits {
    template <typename T>
    using SmartPointerClass = SP<T>;
};

AttributeTraits 使用别名来保存它拥有的模板类。现在，我需要获取特征类的属性类将使用智能指针类的类型定义/别名，例如:

template <typename Traits>
struct Attribute {
    typedef typename Traits::SmartPointerClass<int> SmartPointerType; // error: non-template 'SmartPointerClass' used as template
    using SmartPointerType2 = typename Traits::SmartPointerClass<int>; // error: expected ';' before '<' token
    using SmartPointerType3 = Traits::SmartPointerClass<int>; // error: non-template 'SmartPointerClass' used as template
};

但是上面的 3 个选项都不起作用...

Answer 1

这是你的依赖类是模板的时候之一，所以你必须在定义你的别名时使用笨拙的 template 语法:

template <typename Traits>
struct Attribute {
    typedef typename Traits::template SmartPointerClass<int> SmartPointerType;
    using SmartPointerType2 = typename Traits::template SmartPointerClass<int>; 
    using SmartPointerType3 = typename Traits::template SmartPointerClass<int>; 
};

参见 this documentation有关何时需要 template 关键字的说明:

The template disambiguator for dependent names
Similarly, in a template definition, a dependent name that is not a member of the current instantiation is not considered to be a template name unless the disambiguation keyword template is used or unless it was already established as a template name

This StackOverflow Answer也提供了一些很好的背景。