移动初始化列表的元素是否安全？ [英] Is it safe to move elements of a initializer list?

问题描述

可能重复：

initializer_list和移动语义

在此代码中：

#include <vector>
#include <initializer_list>

template<typename T>
class some_custom_container : public std::vector<T>
{
public:
   some_custom_container(const std::initializer_list<T>& contents)
   {
      for (auto& i : contents)
        this->emplace_back(std::move(i));
   }
};

class test_class
{};

int main()
{
    test_class a;

    some_custom_container<test_class> i = { a, test_class(), a };
}

如果我理解， {a，test_class（），a} 是安全构造的：命名对象被复制，未命名的对象被移动来构造initializer_list。之后，此 initializer_list 通过引用 some_custom_container 的构造函数传递。

If I've understand it, all objects in { a, test_class(), a } are safe-constructed: the named-objects are copied and the unnamed objects are moved to construct the initializer_list. After, this initializer_list is passed by reference to the some_custom_container's constructor.

然后，为了避免无用的doble-copy，我移动所有它们填充向量。

Then, to avoid useless doble-copies I move all of them to fill the vector.

这个构造函数是否安全？我的意思是，在一个奇怪的情况下，例如如果T被评估为参考&

Is it safe this constructor? I mean, in a strange situation, for example if T is evaluated as a reference & or &&, is the vector always well-filled (contains it safe objects)?

如果是这种情况，为什么 initializer_list stl容器的构造函数实现没有以这种方式实现？

If this is the case, why the initializer_list constructor implementations of stl containers aren't implemented in this way? As I know, their constructors copy and don't move the contents.

推荐答案

initializer_list 只提供 const 访问其元素。你可以使用 const_cast 来编译代码，但是移动可能会导致未定义的行为（如果 initializer_list 是真正的const）。所以，没有不安全做这个移动。有这是的解决方法，如果你真的需要它。

initializer_list only provides const access to its elements. You could use const_cast to make that code compile, but then the moves might end up with undefined behaviour (if the elements of the initializer_list are truly const). So, no it is not safe to do this moving. There are workarounds for this, if you truly need it.

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