在“模板”和函数声明之间使用: template<typename trait> Using tr = base_trait<trait> void fn(tr::type arg) { ... }

Question

有没有可能在template关键字和函数声明(以及可能的对象声明等)之间使用别名？

template<typename trait> 
using tr = base_trait<trait> 

void fn(tr::type arg) { // edit: typename has to be here?

} 

如果没有，我该如何实现这种级别的可读性呢？

我想出的变种并不完美：

template<typename trait>
struct base_trait : trait {
    using thetype = typename trait::thetype;
};


struct int_trait {
    using thetype = int;
};


/* the pretty version which doesn't compile */
template <typename trait>
using tr = base_trait<trait>. 

void fn(tr::thetype arg) {
    // tr::

}

fn<int_trait>(1);


/* ugly long arglist version which works */
template <typename trait>
void fn(typename base_trait<trait>::thetype arg1, typename base_trait<trait>::thetype arg2) {
    using tr = base_trait<trait>; 

    // tr:: 
}

fn<int_trait>(1);


/* another boilerplate version that still requires `typename` */
template <typename tr>
void fn(typename tr::thetype arg) {
    // typename tr::  

    // wait, I have to type and read typename everywhere... (unless I add `using boilerplateusing = tr; edit: this using should not help?)
}

template <typename trait>
using boilerplateusing = fn<base_trait<trait>>;

boilerplateusing<int_trait>(1);

Answer 1

只需添加另一个默认带有特征的type成员的模板参数，如下所示

template<typename trait, typename arg_t = typename base_trait<trait>::thetype> 
void fn(arg_t arg) {

} 

或者如果你想保留你可以使用的特征类型

template<typename trait, typename tr = base_trait<trait>> 
void fn(typename tr::thetype arg) {

}

或者您可以使用以下命令同时获得两者

template<typename trait, 
         typename tr = base_trait<trait>,
         typename arg_t = typename tr::thetype> 
void fn(arg_t arg) {

}