如何脱离std :: experimental :: optional? [英] How to disengage std::experimental::optional?

问题描述

借助Boost，我可以使用以下方法创建一个可选的原位:

With Boost I can create an optional in-place with:

boost::optional<boost::asio::io_service::work> work = boost::in_place(boost::ref(io_service));

并通过以下方式将其脱离:

And disengage it with:

work = boost::none;

有了C ++ 14/实验性支持，我可以改为使用以下方法构造一个可选的就地:

With C++14 / experimental support, I can instead construct an optional in-place with:

std::experimental::optional<boost::asio::io_service::work> work;
work.emplace(boost::asio::io_service::work(io_service));

但是我对如何脱离它感到迷茫...

But I'm at a loss for how to disengage it...

推荐答案

work = std::experimental::nullopt;

应脱离work.
库基础TS 在[optional.nullopt ]:

should disengage work.
The library fundamental TS specifies in [optional.nullopt]:

struct nullopt_t是用作唯一的空结构类型 类型，以指示可选对象的脱离状态.

The struct nullopt_t is an empty structure type used as a unique type to indicate a disengaged state for optional objects.

有一个合适的赋值运算符，[optional.object.assign]:

There is an appropriate assignment operator, [optional.object.assign]:

optional<T>& operator=(nullopt_t) noexcept;

效果:如果启用了*this，则调用val->T::~T()销毁所包含的值；否则，调用val->T::~T()销毁所包含的值.否则没有效果.

Effects: If *this is engaged calls val->T::~T() to destroy the contained value; otherwise no effect.

为避免每次都构造一个nullopt_t对象，已经声明了此类型的常量:

To avoid constructing a nullopt_t object every time, a constant of this type is already declared:

struct nullopt_t{ 请参见下文 };
constexpr nullopt_t nullopt( 未指定 );

struct nullopt_t{see below};
constexpr nullopt_t nullopt(unspecified);

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