'auto'作为函数参数的模板参数占位符

问题描述

限时删除！！

C ++ 20允许将auto用作函数参数类型.

它是否还允许将auto用作模板参数占位符(不相似，但是本着(以某种方式)?

因此，以下代码是C ++ 20之前的版本:

template<typename First, typename Second>
void printPair(const std::pair<First, Second>& p) {
    std::cout << p.first << ", " << p.second;
}

可以写为:

void printPair(const std::pair<auto, auto>& p) {
    std::cout << p.first << ", " << p.second;
}

它可以通过实验性GCC概念实现，并且可以很好地工作. >

在C ++ 20中是合法语法吗?

解决方案

此语法在C ++概念技术规范中有效，但在C ++ 20中无效.在C ++ 20概念中，auto仅在函数参数类型的顶层被允许.相关规则是 [dcl.spec.auto]第2段:

(如果您在撰写本文时检查了最新工作草案中的措辞，您会发现一条稍有不同的规则.以上规则已由核心问题2447 ，一周前在布拉格委员会会议上被投票通过了C ++ 20最终草案.)

函数参数中的 decl-specifier 是参数声明开始处的关键字和类型名称的初始序列.上面的规则允许auto出现在顶层:

void f(auto x);

...，但仅作为 decl-specifier .当嵌套在 decl-specifier 中时，不允许auto:

void f(std::vector<auto> x);

...，并且在参数类型的其他地方也不允许使用

void f(void (*p)(auto));

C++20 allows using auto for function parameter type.

Does it also allow using auto as a template argument placeholder (not similar, but in the spirit of C++17 template<auto> in a way) for function parameter type?

So the following code, pre C++20:

template<typename First, typename Second>
void printPair(const std::pair<First, Second>& p) {
    std::cout << p.first << ", " << p.second;
}

Could be written as:

void printPair(const std::pair<auto, auto>& p) {
    std::cout << p.first << ", " << p.second;
}

It does compile and works nicely with experimental GCC implementation for concepts.

Is it a legitimate syntax with C++20?

解决方案

This syntax is valid in the C++ Concepts Technical Specification, but not in C++20. In C++20 concepts, auto is only permitted at the top level in a function parameter type. The relevant rule is [dcl.spec.auto] paragraph 2:

(If you check the wording in the most recent working draft at the time of writing, you will find a somewhat different rule. The above rule was modified by core issue 2447, which was voted into the C++20 final draft at the Prague committee meeting a week ago.)

The decl-specifiers in a function parameter are the initial sequence of keywords and type names at the start of the parameter declaration. The above rule allows auto there at the top level:

void f(auto x);

... but only as a decl-specifier. auto is not permitted when nested within a decl-specifier:

void f(std::vector<auto> x);

... and is also not permitted elsewhere in the parameter type:

void f(void (*p)(auto));

这篇关于'auto'作为函数参数的模板参数占位符的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！

1403页，肝出来的..