c++ - 为什么move构造函数不通过using声明继承

Question

在下面的代码中，虽然基类是可移动构造的，但显然没有生成派生类的移动构造函数。

#include <cstddef>
#include <memory>
#include <cstring>
#include <cassert>

template <typename T>
class unique_array : public std::unique_ptr<T[],void (*)(void*)>
{   size_t Size;
 protected:
    typedef std::unique_ptr<T[],void (*)(void*)> base;
    unique_array(T* ptr, size_t size, void (*deleter)(void*)) noexcept : base(ptr, deleter), Size(size) {}
 public:
    constexpr unique_array() noexcept : base(NULL, operator delete[]), Size(0) {}
    explicit unique_array(size_t size) : base(new T[size], operator delete[]), Size(size) {}
    unique_array(unique_array<T>&& r) : base(move(r)), Size(r.Size) { r.Size = 0; }
    void reset(size_t size = 0) { base::reset(size ? new T[size] : NULL); Size = size; }
    void swap(unique_array<T>&& other) noexcept { base::swap(other); std::swap(Size, other.Size); }
    size_t size() const noexcept { return Size; }
    T* begin() const noexcept { return base::get(); }
    T* end() const noexcept { return begin() + Size; }
    T& operator[](size_t i) const { assert(i < Size); return base::operator[](i); }
    unique_array<T> slice(size_t start, size_t count) const noexcept
    {   assert(start + count <= Size); return unique_array<T>(begin() + start, count, [](void*){}); }
};

template <typename T>
class unique_num_array : public unique_array<T>
{   static_assert(std::is_arithmetic<T>::value, "T must be arithmetic");
 public:
    using unique_array<T>::unique_array;
    unique_num_array(unique_num_array<T>&& r) : unique_array<T>(move(r)) {}
    unique_num_array<T> slice(size_t start, size_t count) const noexcept
    {   assert(start + count <= this->size()); return unique_num_array<T>(this->begin() + start, count, [](void*){}); }
 public: // math operations
    void clear() const { std::memset(this->begin(), 0, this->size() * sizeof(T)); }
    const unique_num_array<T>& operator =(const unique_num_array<T>& r) const { assert(this->size() == r.size()); memcpy(this->begin(), r.begin(), this->size() * sizeof(T)); return *this; }
    const unique_num_array<T>& operator +=(const unique_num_array<T>& r) const;
    // ...
};

int main()
{   // works
    unique_array<int> array1(7);
    unique_array<int> part1 = array1.slice(1,3);
    // does not work
    unique_num_array<int> array2(7);
    unique_num_array<int> part2 = array2.slice(1,3);
    // test for default constructor
    unique_num_array<int> array3;
    return 0;
}

使用上面的代码我得到一个错误(gcc 4.8.4):

test6.cpp: In function ‘int main()’: test6.cpp:47:48: error: use of deleted function ‘unique_num_array::unique_num_array(const unique_num_array&)’ unique_num_array part2 = array2.slice(1,3);

派生类中的切片函数无法按值返回，因为缺少移动构造函数。所有其他构造函数似乎都按预期工作(本示例未涵盖)。

如果我显式定义移动构造函数(取消注释行)，该示例将编译。但在这种情况下，默认构造函数消失了，这当然不是故意的。

这是怎么回事？我不明白这两种情况。

为什么在第一种情况下删除了移动构造函数？

为什么在第二种情况下会丢弃默认构造函数？其他人似乎幸存下来。

Answer 1

Why is the move constructor deleted in the first case?

因为unique_num_array<T>中有一个用户声明的复制赋值运算符，编译器没有隐式声明任何移动构造函数。 [class.copy.ctor]/8 中的标准说

If the definition of a class X does not explicitly declare a move constructor, a non-explicit 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.

---

Why is the default constructor dropped in the second case?

因为unique_num_array<T>中有一个用户声明的移动构造函数，编译器没有隐式声明默认构造函数。 [class.ctor]/4 中的标准说

... If there is no user-declared constructor for class X, a non-explicit constructor having no parameters is implicitly declared as defaulted ([dcl.fct.def]).

---

此外，由于保证 copy elision，此代码将在 C++17 之后工作.详细地说，在 C++17 之前，上下文的语义

return unique_num_array<T>(...);

和

unique_num_array<int> part2 = array2.slice(1,3);

需要复制/移动操作，而在 C++17 之后，语义变为目标对象由纯右值初始化器初始化，而无需具体化临时对象，因此不需要复制/移动。