错误：调用'std :: __ 1 :: unique_ptr< A，std :: __ 1 :: default_delete< A>隐式删除的副本构造函数>'
我正在构造一个对象，该对象采用 std :: vector< std :: unique_ptr< A> > 作为参数。构造函数的定义如下：
I'm constructing an object that takes a std::vector<std::unique_ptr<A> > as an argument. The constructor is defined like this
class B {
std::vector <std::unique_ptr<A> > e_;
public:
B(std::vector <std::unique_ptr<A> > e) : e_(std::move(e)){}
};
然后用作
std::vector <std::unique_ptr<A> > e;
B b(e);
并且Xcode出现错误
and Xcode presents the error
error: call to implicitly-deleted copy constructor of 'std::__1::unique_ptr<A, std::__1::default_delete<A> >'
:new((void*)__p) _Up(_VSTD::forward<_Args>(__args)...);
^ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~`
即使我使用 std :: move（），为什么错误仍然持续存在？
Why is the error still persisting even though i am using std::move()?
编辑：如果我使用 B b（std :: move（e））而不是 B b（e）），有什么方法可以将 move 逻辑移动到该函数的实现中？
the error seems to vanish if i use B b(std::move(e)) instead of B b(e)), is there any way to move the move logic to the implementation of the function?
您的构造函数参数将通过值传递，这将进行复制，但是您无法复制std :: unique_ptr。通过引用传递应该起作用：
Your constructor argument is pass by value which will make a copy, but you cannot copy a std::unique_ptr. Passing by reference should work:
class B {
std::vector <std::unique_ptr<float> > e_;
public:
B(std::vector <std::unique_ptr<float> >& e) : e_(std::move(e)){}
};
但是...我同意其他评论，认为这是不好的设计。如果您想让 B 拥有 e 但还想操纵 e 在 B 之外，那么它应该是公共成员，不需要花哨的构造函数：
But...I agree with the other comments that this is bad design. If you want B to own e but also want to manipulate e outside of B then it should be a public member, no fancy constructor needed:
class B {
public:
std::vector <std::unique_ptr<float> > e_;
};