为什么由析构函数构造的两个对象被调用三次

Question

这是我对一些C++ 11研究示例的实现。我让所有的构造函数和析构函数打印到控制台。但令人惊讶的是，我得到了构造函数两次，但是析构函数被调用了三次。

似乎意想不到的是0x7fff5fbff6d0。当这个对象被创建时？但为什么没有构造函数调用相关联？

为什么会发生这种情况？

template<typename T> 
class ArrayWrapper{ 
public: 
    ArrayWrapper():data_(nullptr), size_(0){ 
     cout << "Default ctor called "<< this <<endl; 
    } 

    ArrayWrapper(size_t n, const T& val) : data_(new T[n]), size_(n){ 
     cout << "ctor_n_val called "<< this << endl; 
     for_each(data_, data_+size_, [&](T& elem){ elem=val; }); 
    } 

    ArrayWrapper(const ArrayWrapper& other): data_(new T[other.size_]), size_(other.size_) 
    { 
     cout << "copy ctor called "<< this <<endl; 
     copy(other.data_, other.data_+other.size_, data_); 
    } 

    ArrayWrapper(ArrayWrapper&& other): data_(other.data_), size_(other.size_) 
    { 
     cout << "move ctor called"<<endl; 
     other.data_ = nullptr; 
     other.size_ = 0; 
    } 

    ArrayWrapper<T>& operator=(const ArrayWrapper& other){ 
     cout << "copy assignment called" <<endl; 
     if(this != &other){ 
      delete data_; 
      data_ = new T[other.size_]; 
      copy(other.begin(), other.end(), begin()); 
      size_ = other.size_; 
     } 
     return *this; 
    } 

    ArrayWrapper<T> operator=(ArrayWrapper&& other){ 
     cout << "move assignment called " <<this << " <- " <<&other <<endl; 
     swap(size_, other.size_); 
     swap(data_, other.data_); 
    } 

    ~ArrayWrapper(){ 
     cout <<"Destroying " << this << " Size " << size_ <<endl; 
    } 
    typedef T* iterator; 
    typedef const T* const_iterator; 

    T* begin() { 
     return data_; 
    } 

    T* end(){ 
     return data_ + size_; 
    } 

    const T* begin() const { 
     return data_; 
    } 

    const T* end() const { 
     return data_ + size_; 
    } 

    const T* cbegin() const { 
     return data_; 
    } 

    const T* cend() const { 
     return data_ + size_; 
    } 

    size_t size(){ 
     return size_; 
    } 
public: 
    T* data_; 
    size_t size_; 
}; 

template<typename T> 
ArrayWrapper<T> make_array(size_t n, const T& val){ 
    cout <<"Factory method called"<<endl; 
    return ArrayWrapper<T>(n, val); 
} 

template<typename T> 
std::ostream& operator<<(std::ostream& os, const ArrayWrapper<T>& arr){ 
    for(const T& elem: arr){ os << elem << ", ";} 
    return os; 
} 

int main(){ 
    size_t n = 10; 
    ArrayWrapper<int> a4(n, 1); 
    a4 = make_array(n, 4); // move assignment: 
    cout << "A4: " << a4 << endl; 
} 

输出：

$ g++-mp-4.8 -std=c++11 move.cpp 

$ ./a.out 

ctor_n_val called 0x7fff5fbff6b0 

Factory method called 

ctor_n_val called 0x7fff5fbff6e0 

move assignment called 0x7fff5fbff6b0 <- 0x7fff5fbff6e0 

Destroying 0x7fff5fbff6d0 Size 0 

Destroying 0x7fff5fbff6e0 Size 10 

A4: 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 

Destroying 0x7fff5fbff6b0 Size 10 

Answer 1

你我的行动赋值运算符应该返回一个参考：

ArrayWrapper<T>& operator=(ArrayWrapper&& other) 
//   ^

既然你有它的价值回归，但目前还没有return说法，你调用未定义的行为。你应该实现它像拷贝赋值运算符，除了当然移动的资源，而不是将它们复制：

ArrayWrapper<T>& operator=(ArrayWrapper&& other){ 
    if(this != &other){ 
     delete[] data_; 
     size_ = other.size_; 
     data_ = other.data_; 
     other.size_ = 0; 
     other.data_ = nullptr; 
    } 
    return *this; 
} 

另外，还要注意使用delete[]删除动态分配数组。