为什么矢量 <T> :: \\ templace_back，这样T有一个删除的拷贝构造函数，无法编译？

Question

I cannot compile the following dont_compile function. I don't understand why it doesn't work. But, it does work with list .

class Thing {
public:
    Thing() {}
    Thing(const Thing &) = delete;
};

int dont_compile(int argc, char ** argv)
{
    std::vector<Thing> v;
    v.emplace_back();

    return 0;
}

int compiles(int argc, char ** argv)
{
    std::list<Thing> v;
    v.emplace_back();

    return 0;
}

Here is the error from the compiler. Is it a bug?

/Applications/Xcode.app/Contents/Developer/Toolchains/XcodeDefault.xctoolchain/usr/bin/../include/c++/v1/memory:1752:31: error: call to deleted constructor of 'Thing'
            ::new((void*)__p) _Up(_VSTD::forward<_Args>(__args)...);
                              ^   ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

... snip ...

note: 'Thing' has been explicitly marked deleted here
        Thing(const Thing &) = delete;

I don't really understand how _Up(...) is causing the copy constructor to be invoked.

Answer 1

std::vector::emplace_back requires the type of the vector to be EmplaceConstructible as well as MoveInsertable . Since you delete the copy constructor and do not define a move constructor yourself, Thing does not satisfy the second requirement. In contrast, std::list::emplace_back only requires the list type to be EmplaceConstructible .

Answer 2

It works when you have move constructor:

#include <vector>

class Thing {
public:
    Thing() {}
    Thing(const Thing &) = delete;
    Thing(Thing&&) = default;
};

int main() {
    std::vector<Thing> v;
    v.emplace_back();
    return 0;
}

The type requirements of std::vector::emplace_back can provide more information.