它是可以移动一个boost ::可选？ [英] Is it possible to move a boost::optional?

问题描述

我一直试图在一个类中定义一个默认的构造函数的举动与的boost ::可选的成员变量。

 的#include＆LT;升压/ optional.hpp＆GT;
＃包括LT＆;实用＆GT;
＃包括LT＆;矢量＆GT;结构bar {性病::矢量＆lt;＆INT GT; VEC;};结构foo的{
  富（）=默认值;
  富（富和放大器;＆安培）=默认值;
  推动::可选＆LT; BAR＆GT;你好;
};诠释主（）{
  foo的;
  富B（的std ::移动（一））;
}
 

我的编译器同时支持移动语义和违约举动构造函数，但我不能得到这个工作。

 ％铛++ foo.cc -std = C ++ 11 -stdlib = ++的libc
foo.cc:15:7：错误：调用'富'的构造函数中删除
  富B（的std ::移动（一））;
      ^ ~~~~~~~~~~~~
foo.cc:9:3：注意：功能已被明确标记为删除这里
  富（富和放大器;＆安培）=默认值;
  ^
产生1个错误。
 

有没有移动的boost ::可选的方式的没有的修改Boost的源$ C ​​$ C？或者我应该等到加速支持移动？

我有同样的问题，的boost ::任何过去。

解决方案

看着的boost ::可选的源$ C ​​$ C，它没有定义移动构造或移动作业。但是，它定义拷贝构造函数和赋值，其中prevent从自动生成移动构造函数和赋值编译器吧。

请参阅移动构造 - Q＆安培; A ：

  

编译器会生成隐式转移构造函数作为成员明智之举，除非你有明确定义一个拷贝构造函数或复制/移动分配或析构函数

由于吕克·丹东表明需要有针对治疗。让我们尝试使用交换移动的boost ::可选的，希望创建一个空的提振::可选的便宜，然后我们只要将默认的构造的boost ::可选的从R值<$ C $之一C>富，例如：

 富（富和放大器;和b）{
    hello.swap（b.hello）;
}
 

和同为移动作业。

I've been trying to define a defaulted move constructor in a class with a boost::optional member variable.

#include <boost/optional.hpp>
#include <utility>
#include <vector>

struct bar {std::vector<int> vec;};

struct foo {
  foo() = default;
  foo(foo&&) = default;
  boost::optional<bar> hello;
};

int main() {
  foo a;
  foo b(std::move(a));
}

My compiler supports both move semantics and defaulted move constructors, but I cannot get this to work.

% clang++ foo.cc -std=c++11 -stdlib=libc++
foo.cc:15:7: error: call to deleted constructor of 'foo'
  foo b(std::move(a));
      ^ ~~~~~~~~~~~~
foo.cc:9:3: note: function has been explicitly marked deleted here
  foo(foo&&) = default;
  ^
1 error generated.

Is there a way to move a boost::optional without modifying Boost's source code? Or should I wait until Boost supports moving?

I had the same problem with boost::any in the past.

解决方案

Looking at boost::optional source code, it doesn't define move constructors or move assignments. However, it does define copy constructor and assignment, which prevent the compiler from auto-generating move constructor and assignment for it.

See Move constructor — Q&A:

the compiler will implicitly generate move constructor as member-wise moves, unless you have explicitly defined a copy constructor or copy/move assignment or a destructor

As Luc Danton suggests there needs to be a cure for that. Let's try using swap to move boost::optional in the hope that creating an empty boost::optional is cheap and then we just swap the default-constructed boost::optional with the one from the r-value foo, e.g.:

foo(foo&& b) {
    hello.swap(b.hello);
}

And same for the move assignment.

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