[英]Why is copy constructor called in call to std::vector::emplace_back()?
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. 我的理解是
std::vector::emplace_back()
的目的是专门避免调用复制构造函数,而是直接构造对象。
Consider the following code: 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());
}
For some reason, despite the use of the emplace_back()
function, this code fails to compile, with the error: 出于某种原因,尽管使用了
emplace_back()
函数,但此代码无法编译,错误如下:
error C2248: 'std::unique_ptr<_Ty>::unique_ptr' : cannot access private member declared in class 'std::unique_ptr<_Ty>' [...] This diagnostic occurred in the compiler generated function 'stuff::stuff(const stuff &)'
Notice that the compiler attempted to create (and use) the COPY CONSTRUCTOR . 请注意,编译器尝试创建(和使用)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. 正如我上面所讨论的,我的理解是
emplace_back()
的目的是避免使用复制构造函数。
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. 当然,由于编译器试图创建并调用复制构造函数, 即使我为
stuff
定义了复制构造函数,代码也无法编译,因为std::unique_ptr
不能在复制构造函数中使用。 Hence, I would very much like to avoid the use of a copy constructor (in fact, I need to avoid it). 因此,我非常希望避免使用复制构造函数(事实上,我需要避免使用它)。
(This is VS 11.0.60610.01 Update 3 on Windows 7 64-bit) (这是Windows 7 64位上的VS 11.0.60610.01 Update 3)
Why is the compiler generating, and attempting to use, the copy constructor, even though I am calling emplace_back()
? 为什么编译器会生成并尝试使用复制构造函数,即使我正在调用
emplace_back()
?
Note (in response to @Yakk's answer): 注意(回应@ Yakk的回答):
Explicitly adding the move constructor, as follows, resolves the problem: 显式添加移动构造函数,如下所示,可以解决问题:
stuff(stuff && rhs)
: dummy_ptr(std::move(rhs.dummy_ptr))
, dummy_path(rhs.dummy_path)
{}
Visual Studio 2013 and earlier fails to write default move constructors for you. Visual Studio 2013及更早版本无法为您编写默认移动构造函数。 Add a simple explicit move constructor to
stuff
. 一个简单明确的移动构造函数添加到
stuff
。
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. 如果需要重新分配,推送或安装后退可以导致移动东西,在你的情况下复制,因为
stuff
没有移动。
It is a msvc bug. 这是一个msvc错误。
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