什么是模板构造函数继承的标准符合语法？ [英] What is the standard conform syntax for template constructor inheritance?

问题描述

GCC 4.8.1接受

GCC 4.8.1 accepts

template <typename T>
class Subclass : public Baseclass<T>
{
public:
    using typename Baseclass<T>::Baseclass;
};

但MSVC没有。另一方面，MSVC接受

but MSVC does not. On the other hand, MSVC accepts

template <typename T>
class Subclass : public Baseclass<T>
{
public:
    using typename Baseclass::Baseclass;
};

但GCC没有。然后，我在这个问题中看到了另一种声明： c ++ 11继承模板构造函数

but GCC does not. Then I've seen another kind of declaration in this questions: c++11 inheriting template constructors

template <typename T>
class Subclass : public Baseclass<T>
{
public:
    using typename Baseclass::Baseclass<T>;
};

b

for which MSVC warns about an "obsolete declaration style" and GCC says

prog.cpp:8:24: error: ‘template<class T> class Baseclass’ used without template parameters
        using typename Baseclass::Baseclass<T>;

我认为第一个例子是标准的符合语法。直观地，它看起来很对我。

I thought the first example would be the standard conform syntax. Intuitively, it looks right to me.

什么是c ++ 11标准符合语法？

What is the c++11 standard conform syntax?

推荐答案

答案有点埋在标准中。使用声明定义为（7.3.3）：

The answer is a bit buried in the standard. A using declaration is defined as (7.3.3):

using [typename] nested-name-specifier unqualified-id;

nested-name-specifier 一些步骤转换为 simple-template-id （定义为

The nested-name-specifier resolves after some steps into simple-template-id which is defined as

template-name < [template-argument-list] >

简而言之，符合标准的语法是

In short, the standard conforming syntax is

template <typename T>
class Subclass : public Baseclass<T>
{
public:
    using typename Baseclass<T>::Baseclass;
};

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