c++ - 为什么具有用户声明的析构函数的类具有隐式默认的构造函数？

Question

代码:

struct A
{ 
    ~A(){ };
};

A::A(){ }; //error: definition of implicitly declared default constructor

int main()
{
    A a;
}

DEMO

为什么代码会产生错误？我希望程序编译正常。标准说 N3797::12.8/7 [class.copy]:

If the class definition does not explicitly declare a copy constructor, one is declared implicitly. If the class definition declares a move constructor or move assignment operator, the implicitly declared copy constructor is defined as deleted; otherwise, it is defined as defaulted (8.4). The latter case is deprecated if the class has a user-declared copy assignment operator or a user-declared destructor.

是bug还是我误会了？

Answer 1

struct A
{ 
  ~A(){ };
   A();
};

A::A(){ }; //here you can define default constructor

int main()
{
 A a;
}

你已经定义了显式析构函数而不是构造函数，添加构造函数声明并将其定义为外部类