在繁重的模板元编程上下文中考虑基础模板类和派生模板类(此处为了易于阅读并着重于此问题而对其进行了简化)。
template <class T>
struct base {
using type = T;
static constexpr int value = 0;
template <class... Args>
constexpr void function(Args&&...) {}
template <class U>
using alias_template = base<U>;
template <class U>
static constexpr int variable_template = 0;
};
template <class T>
struct derived: base<T> {
using typename base<T>::type; // Using base typedef
using base<T>::value; // Using base static data member
using base<T>::function; // Using base function members (regardless of template or not)
//using typename base<T>::alias_template; // DOES NOT SEEM TO WORK
//using base<T>::variable_template; // DOES NOT SEEM TO WORK
using typedef_test = type; // Working
static constexpr int value_test = value; // Working
using alias_template_test = alias_template<T>; // ERROR
static constexpr int variable_template_test = variable_template<T>; // ERROR
};
问题:是否存在
using
语法来公开从基类继承的别名模板和变量模板,以便编译当前错误的行?有什么解决方法可以避免在派生类中每次都指定base<T>::
(这里仍然很简单,但是在我的实际代码中,每次指定都很快变得很烦人)?
最佳答案
using
doesn't work this way用于依赖成员模板:
A using-declaration also can't be used to introduce the name of a dependent member template as a template-name (the template disambiguator for dependent names is not permitted)
我不确定这是否满足您的要求,但是为了使别名像模板一样工作,您可以在根据基本成员模板定义的派生类中声明新的成员模板:
template<typename U>
using alias_template = typename base<T>::template alias_template<U>;
template<typename U>
static constexpr auto variable_template = base<T>::template variable_template<U>;
但是,IMO每次指定
base<T>::
都不是问题,并且比引入新模板更干净。您甚至可以使用using Base = my_long_base_class_name<T>;
将其缩短
关于c++ - 使用语法在派生类中公开基类别名模板和变量模板?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/61857361/