为什么在对std :: vector :: emplace_back（）的调用中调用了复制构造函数？ [英] Why is copy constructor called in call to std::vector::emplace_back()?

问题描述

据我了解， std :: vector :: emplace_back（）的目的是专门避免调用复制构造函数，并且

It is my understanding that the purpose of std::vector::emplace_back() is specifically to avoid calling a copy constructor, and instead to construct the object directly.

请考虑以下代码：

#include <memory>
#include <vector>
#include <boost/filesystem.hpp>

using namespace std;

struct stuff
{
    unique_ptr<int> dummy_ptr;
    boost::filesystem::path dummy_path;
    stuff(unique_ptr<int> && dummy_ptr_,
          boost::filesystem::path const & dummy_path_)
        : dummy_ptr(std::move(dummy_ptr_))
        , dummy_path(dummy_path_)
    {}
};

int main(int argc, const char * argv[])
{

    vector<stuff> myvec;

    // Do not pass an object of type "stuff" to the "emplace_back()" function.
    // ... Instead, pass **arguments** that would be passed
    // ... to "stuff"'s constructor,
    // ... and expect the "stuff" object to be constructed directly in-place,
    // ... using the constructor that takes those arguments
    myvec.emplace_back(unique_ptr<int>(new int(12)), boost::filesystem::path());

}

由于某些原因，尽管使用了 emplace_back（）函数，此代码无法编译，并显示错误：

For some reason, despite the use of the emplace_back() function, this code fails to compile, with the error:

错误C2248：' std :: unique_ptr< _Ty> :: unique_ptr'：无法访问在类'std :: unique_ptr< _Ty>'中声明的私有成员[...]此诊断发生在编译器生成的函数'stuff :: stuff（const stuff& amp; ;）'

请注意，编译器试图创建（和使用）COPY CONSTRUCTOR 。如上所述，据我了解， emplace_back（）的目的是避免使用复制副本的构造函数。

Notice that the compiler attempted to create (and use) the COPY CONSTRUCTOR. As I've discussed above, it's my understanding that the purpose of emplace_back() is to avoid the use of the copy constructor.

当然，由于编译器试图创建并调用 copy 构造函数，因此代码无法编译即使我为东西定义了复制构造函数，因为 std :: unique_ptr 不能为在复制构造函数中使用。因此，我非常想避免使用复制构造函数（实际上，我需要避免使用它）。

Of course, since the compiler is attempting to create and call the copy constructor, there's no way the code would compile even if I defined the copy constructor for stuff, because the std::unique_ptr cannot be used in a copy constructor. Hence, I would very much like to avoid the use of a copy constructor (in fact, I need to avoid it).

（这是VS 11.0.60610.01 Update 3在Windows 7 64位系统上）

(This is VS 11.0.60610.01 Update 3 on Windows 7 64-bit)

为什么编译器会生成并尝试使用副本构造函数，即使我正在调用 emplace_back（ ）？

Why is the compiler generating, and attempting to use, the copy constructor, even though I am calling emplace_back()?

注意（响应@Yakk的回答）：

Note (in response to @Yakk's answer):

按如下所示显式添加move构造函数可以解决问题：

Explicitly adding the move constructor, as follows, resolves the problem:

stuff(stuff && rhs)
    : dummy_ptr(std::move(rhs.dummy_ptr))
    , dummy_path(rhs.dummy_path)
{}

推荐答案

Visual Studio 2013及更早版本无法为您编写默认的move构造函数。将简单的显式move构造函数添加到 stuff 。

Visual Studio 2013 and earlier fails to write default move constructors for you. Add a simple explicit move constructor to stuff.

如果将其推入或放回原位，则可能导致其移动需要重新分配，这在您的情况下是复制的，因为东西没有任何动作。

A push or emplace back can cause stuff to be moved if it needs to reallocate, which in your case copies, as stuff has no move.

这是msvc错误。

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