具有显式析构函数和 std::unique_ptr<> 成员的 Class 不能在 std::vector<> 中使用？

Question

这个代码

#include <memory>
#include <vector>

class Foo
{
public:
    ~Foo()
    {
    }

    std::unique_ptr<int> bar;
};

int main()
{
    std::vector<Foo> foos;
    foos.emplace_back();
}

在 g++ 中产生以下错误消息：

In file included from /usr/include/c++/4.8/memory:64:0,
                 from main.cpp:1:
/usr/include/c++/4.8/bits/stl_construct.h: In instantiation of ‘void std::_Construct(_T1*, _Args&& ...) [with _T1 = Foo; _Args = {Foo}]’:
/usr/include/c++/4.8/bits/stl_uninitialized.h:75:53:   required from ‘static _ForwardIterator std::__uninitialized_copy<_TrivialValueTypes>::__uninit_copy(_InputIterator, _InputIterator, _ForwardIterator) [with _InputIterator = std::move_iterator<Foo*>; _ForwardIterator = Foo*; bool _TrivialValueTypes = false]’
/usr/include/c++/4.8/bits/stl_uninitialized.h:117:41:   required from ‘_ForwardIterator std::uninitialized_copy(_InputIterator, _InputIterator, _ForwardIterator) [with _InputIterator = std::move_iterator<Foo*>; _ForwardIterator = Foo*]’
/usr/include/c++/4.8/bits/stl_uninitialized.h:258:63:   required from ‘_ForwardIterator std::__uninitialized_copy_a(_InputIterator, _InputIterator, _ForwardIterator, std::allocator<_Tp>&) [with _InputIterator = std::move_iterator<Foo*>; _ForwardIterator = Foo*; _Tp = Foo]’
/usr/include/c++/4.8/bits/stl_uninitialized.h:281:69:   required from ‘_ForwardIterator std::__uninitialized_move_if_noexcept_a(_InputIterator, _InputIterator, _ForwardIterator, _Allocator&) [with _InputIterator = Foo*; _ForwardIterator = Foo*; _Allocator = std::allocator<Foo>]’
/usr/include/c++/4.8/bits/vector.tcc:415:43:   required from ‘void std::vector<_Tp, _Alloc>::_M_emplace_back_aux(_Args&& ...) [with _Args = {}; _Tp = Foo; _Alloc = std::allocator<Foo>]’
/usr/include/c++/4.8/bits/vector.tcc:101:54:   required from ‘void std::vector<_Tp, _Alloc>::emplace_back(_Args&& ...) [with _Args = {}; _Tp = Foo; _Alloc = std::allocator<Foo>]’
main.cpp:17:23:   required from here
/usr/include/c++/4.8/bits/stl_construct.h:75:7: error: use of deleted function ‘Foo::Foo(const Foo&)’
     { ::new(static_cast<void*>(__p)) _T1(std::forward<_Args>(__args)...); }
       ^
main.cpp:4:7: note: ‘Foo::Foo(const Foo&)’ is implicitly deleted because the default definition would be ill-formed:
 class Foo
       ^
main.cpp:4:7: error: use of deleted function ‘std::unique_ptr<_Tp, _Dp>::unique_ptr(const std::unique_ptr<_Tp, _Dp>&) [with _Tp = int; _Dp = std::default_delete<int>]’
In file included from /usr/include/c++/4.8/memory:81:0,
                 from main.cpp:1:
/usr/include/c++/4.8/bits/unique_ptr.h:273:7: error: declared here
       unique_ptr(const unique_ptr&) = delete;

然而，如果我删除 ~Foo() 实现（和声明），它编译得很好。可以观察到相同的行为 main() 如下所示：

int main()
{
    auto f = Foo();
}

为什么 Foo 的复制构造函数仍然被调用？这里的一切不应该通过移动语义发生吗？
如果我删除析构函数，为什么它（复制构造函数）不会被隐式删除？

Answer 1

显式析构函数声明（和定义）抑制了移动构造函数和移动赋值运算符的隐式声明。因此，唯一可以使用的是复制构造函数（无论如何都隐式声明） - 但它被定义为已删除，因为 unique_ptrs 不可复制。因此错误。

显式默认它们 - 以及默认构造函数，因为您现在有一个用户声明的构造函数：

Foo() = default;
Foo(Foo&&) = default;
Foo& operator=(Foo&&) = default;

Why wouldn't it (the copy constructor) be implicitly deleted if I remove the destructor?

它仍然被隐式删除。但是由于现在隐式声明了一个移动构造函数，移动构造函数可以用于移动，而复制构造函数根本就不用了。

相关标准引述（§12.8 [class.copy]/p9, 20, 强调已加）：

9 If the definition of a class X does not explicitly declare a move constructor, one will be implicitly declared as defaulted if and only if

X does not have a user-declared copy constructor,

X does not have a user-declared copy assignment operator,

X does not have a user-declared move assignment operator, and

X does not have a user-declared destructor.

20 If the definition of a class X does not explicitly declare a move assignment operator, one will be implicitly declared as defaulted if and only if

X does not have a user-declared copy constructor,

X does not have a user-declared move constructor,

X does not have a user-declared copy assignment operator, and

X does not have a user-declared destructor.

Answer 2

Foo 不可复制，因为它有一个成员 unique_ptr。 Foo 也不是 movable 因为用户提供的析构函数的声明抑制了默认移动构造函数/赋值的隐式生成。如果你希望 Foo 是可移动的默认移动 constructor/assignment 所以编译器会为你生成它们：

class Foo
{
public:
    ~Foo()
    {
    }

    Foo(Foo&&) = default;
    Foo& operator=(Foo&&) = default;

    std::unique_ptr<int> bar;
};

具有显式析构函数和 std::unique_ptr<> 成员的 Class 不能在 std::vector<> 中使用？

Class with explicit destructor and std::unique_ptr<> member can't be used in std::vector<>?

c++

c++11

move-semantics