检查后的防护参数包是否会导致程序格式错误? [英] Are checked guard parameter packs cause of ill-formed programs?

问题描述

不止一次(甚至在SO上)，我也看到过这样的代码:

More than once (even on SO) I've seen code like this:

template<typename U, typename... G, typename T = Traits<U>>
struct {
    static_assert(sizeof...(G) == 0, "!");
    // ...
};

或者这个:

template<typename T, typename... G, typename = std::enable_if_t<condition<T>>
void func(T &&t) {
    static_assert(sizeof...(G) == 0, "!");
    // ....
}

目的是通过执行以下操作来避免用户破坏游戏规则:

The intent was to avoid users break the rule of the game by doing something like this:

template<typename T, typename = std::enable_if_t<std::is_same<T, int>>
void func(T &&t) {
    // ....
}

// ...

func<int&, void>(my_int);

使用保护参数包，不能覆盖默认值.
另一方面，对大小的检查避免了无用参数对专业化的污染.

With the guard parameter pack, the defaulted value cannot be overridden.
On the other side, the check on the size avoids pollution of the specializations with useless parameters.

无论如何，由于 [temp.res/8] ，我们有那个:

Anyway, because of [temp.res/8], we have that:

该程序格式错误，如果满足以下条件，则无需进行诊断:
[...]
-可变参数模板的每个有效专业化都需要一个空的模板参数包，或者
[...]

The program is ill-formed, no diagnostic required, if:
[...]
- every valid specialization of a variadic template requires an empty template parameter pack, or
[...]

因此，包含上述摘录的程序是否格式错误?

Therefore, are the programs that contain the above mentioned snippets ill-formed or not?

推荐答案

技巧"导致程序格式错误，不需要诊断.

The "trick" results in an ill formed program, no diagnostic required.

该标准在您引用的部分中有明确说明.

The standard states it clearly in the section you quoted.

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