格式错误的呼叫的零长度可变参数展开 [英] zero length variadic expansion of ill-formed call

问题描述

标准大师的问题.

尝试回应另一个问题，我开始怀疑代码的格式是否正确.

Trying to respond to another question, I came to doubt about the well-formedness of a code.

据我所知，以下代码格式错误

As far I know, the following code is ill-formed

int main ()
 {
   std::tuple<>  a;

   std::get<0>(a);
 }

由于t是std::tuple<Ts...>时对std::get<I>(t)的调用在I超出[0, sizeof...(Ts)[范围时是错误的.

because a call to std::get<I>(t), when t is a std::tuple<Ts...>, is ill-formed when I is outside the range [0, sizeof...(Ts)[.

在这种情况下，sizeof...(Ts)为零，因此范围[0, 0[为空，因此对于每个索引I，std::get<I>(a)都是错误的.

In this case sizeof...(Ts) is zero, so the range [0, 0[ is empty, so std::get<I>(a) is ill-formed for every index I.

但是当std::get<I>(a)通过一个空的可变参数包展开时?

But when std::get<I>(a) is expanded through an empty variadic pack?

我的意思是:以下代码

#include <tuple>

template <typename ... Args>
void bar (Args const & ...)
 { }

template <std::size_t ... I>
void foo ()
 {
   std::tuple<> a;

   bar( std::get<I>(a) ... );
 }

int main ()
 {
   foo<>();
 }

使用格式错误的(?)调用(std::get<I>(a))但零时间可变参数扩展(sizeof...(I)为零)，格式正确还是格式错误?

that uses a ill-formed (?) call (std::get<I>(a)) but zero-time variadic expanded (sizeof...(I) is zero), is well-formed or ill-formed?

推荐答案

.res]/8 :

该程序格式错误，如果出现以下情况，则无需进行诊断:

The program is ill-formed, no diagnostic required, if:

• [...]
• 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
• [...]

• [...]
• every valid specialization of a variadic template requires an empty template parameter pack, or
• [...]

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