一个类有多少个默认方法？

Question

抱歉，这可能看起来很简单，但有人问过我这个问题，我不确定。

一个空的C++类有什么功能？

构造函数， 复制构造函数， 任务， 析构函数？

是这样吗？还是还有更多？

Answer 1

在C++03中有4个：

• 默认构造函数：仅在以下情况下声明 没有用户定义的构造函数 宣布。使用时定义

• 复制构造函数 - 仅在用户未声明时才声明。如果使用则定义

• 复制赋值运算符同上

• 析构函数同上

在 C++11 中还有两个：

• 移动构造函数
• 移动赋值运算符

编译器也可能无法生成其中的一些。例如，如果类包含例如引用（或任何其他无法复制分配的内容），则编译器将无法为您生成复制分配运算符。 For more information read this

Answer 2

如果我定义以下类

class X
{};

编译器会定义以下方法：

X::X()  {}                    // Default constructor. It takes zero arguments (hence default).
X::~X() {}                    // Destructor
X::X(X const& rhs) {};        // Copy constructor
X& operator=(X const& rhs)
{return *this;}              // Assignment operator.

// Since C++11 we also define move operations
X::X(X&& rhs) {};            // Move Constructor
X& operator=(X&& rhs)
{return *this;}              // Move Assignment

注意：
如果定义了 ANY 构造函数，则不会构建默认构造函数。
如果用户定义了替代方法，则不会构建其他方法。

更有趣的是当我们有成员和基时的默认实现：

class Y: public X
{
    int    a;      // POD data
    int*   b;      // POD (that also happens to be a pointer)
    Z      z;      // A class
};

// Note: There are two variants of the default constructor.
//       Both are used depending on context when the compiler defined version
//       of the default constructor is used.
//
//       One does `default initialization`
//       One does `zero initialization`

// Objects are zero initialized when
//   They are 'static storage duration'
//   **OR** You use the braces when using the default constructor
Y::Y()      // Zero initializer
    : X()   // Zero initializer
    , a(0)
    , b(0)
    , z()   // Zero initializer of Z called.
{}

// Objects are default initialized when
//    They are 'automatic storage duration'
//    **AND** don't use the braces when using the default constructor
Y::Y()
    :X    // Not legal syntax trying to portray default initialization of X (base class)
    //,a  // POD: uninitialized.
    //,b  // POD: uninitialized.
    ,z    // Not legal syntax trying to portray default initialization of z (member)
{}
//
// Note: It is actually hard to correctly zero initialize a 'automatic storage duration'
//       variable (because of the parsing problems it tends to end up a a function
//       declaration). Thus in a function context member variables can have indeterminate
//       values because of default initialization. Thus it is always good practice to 
//       to initialize all members of your class during construction (preferably in the
//       initialization list).
//
// Note: This was defined this way so that the C++ is backward compatible with C.
//       And obeys the rule of don't do more than you need too (because we want the C++
//       code to be as fast and efficient as possible.


Y::Y(Y const& rhs)
    :X(rhs)              // Copy construct the base
    ,a(rhs.a)            // Copy construct each member using the copy constructor.
    ,b(rhs.b)            // NOTE: The order is explicitly defined
    ,z(rhs.z)            //       as the order of declaration in the class.
{}

Y& operator=(Y const& rhs)
{
    X::operator=(rhs);   // Use base copy assignment operator
    a  = rhs.a;          // Use the copy assignment operator on each member.
    b  = rhs.b;          // NOTE: The order is explicitly defined
    z  = rhs.z;          //       as the order of declaration in the class.
    return(*this);
}

Y::~Y()
{
    Your Code first
}
// Not legal code. Trying to show what happens.
  : ~z()
  , ~b() // Does nothing for pointers.
  , ~a() // Does nothing for POD types
  , ~X() ; // Base class destructed last.


// Move semantics:
Y::Y(Y&& rhs)
    :X(std::move(rhs))   // Move construct the base
    ,a(std::move(rhs.a)) // Move construct each member using the copy constructor.
    ,b(std::move(rhs.b)) // NOTE: The order is explicitly defined
    ,z(std::move(rhs.z)) //       as the order of declaration in the class.
{}

Y& operator=(Y&& rhs)
{
    X::operator=(std::move(rhs));   // Use base move assignment operator
    a  = std::move(rhs.a);          // Use the move assignment operator on each member.
    b  = std::move(rhs.b);          // NOTE: The order is explicitly defined
    z  = std::move(rhs.z);          //       as the order of declaration in the class.
    return(*this);
}

Answer 3

只是为了扩展Armen Tsirunyan answer，这里是方法的签名：

// C++03
MyClass();                                     // Default constructor
MyClass(const MyClass& other);                 // Copy constructor
MyClass& operator=(const MyClass& other);      // Copy assignment operator
~MyClass();                                    // Destructor

// C++11 adds two more
MyClass(MyClass&& other) noexcept;             // Move constructor
MyClass& operator=(MyClass&& other) noexcept;  // Move assignment operator

Answer 4

编译器为空类分配的默认方法：

http://cplusplusinterviews.blogspot.sg/2015/04/compiler-default-methods.html

Answer 5

是吗？

是的，就是这样。

编译器默认生成

• 默认构造函数
• 复制构造函数
• 复制赋值运算符
• 析构函数

一堂课

使用Clang的-ast-dump选项可以看到默认生成的构造函数、复制构造函数和赋值操作符

prasoon@prasoon-desktop ~ $ cat empty.cpp && clang++ -cc1 -ast-dump empty.cpp
class empty
{};

int main()
{
   empty e;
   empty e2 = e;
   {
     empty e3;
     e3 = e;
   }

}
typedef char *__builtin_va_list;
class empty {
    class empty;
    inline empty() throw(); //default c-tor
    //copy c-tor
    inline empty(empty const &) throw() (CompoundStmt 0xa0b1050 <empty.cpp:1:7>)

    //assignment operator   
    inline empty &operator=(empty const &) throw() (CompoundStmt 0xa0b1590 <empty.cpp:1:7>
  (ReturnStmt 0xa0b1578 <col:7>
    (UnaryOperator 0xa0b1558 <col:7> 'class empty' prefix '*'
      (CXXThisExpr 0xa0b1538 <col:7> 'class empty *' this))))


};