静态方法class实例化

Question

当默认构造函数为私有时，像这样 Foo t3 = Foo::construct(true); 的 C++ 代码行如何工作？我的假设（这显然是不正确的）是调用默认构造函数，然后是我的赋值运算符。该假设一定是不正确的，因为默认构造函数是私有的，无法调用。

一个具体的例子：

class Foo {
private:
   Foo() {}
   bool bar;
public:
   Foo(bool t): bar(t) {}
   static Foo construct(bool t) {
      Foo temp; //calling private constructor;
      temp.bar = t;
      return temp;
   }
}

实例化此 class 的测试方法如下所示：

int main() {
   //Foo t1; //Not allowed, compile error, Foo() is private
   Foo t2(true); //Constructor, valid use
   Foo t3 = Foo::construct(true); //It works! Why?
   return 0;
}

实例化 t3 时幕后究竟发生了什么？

Answer 1

您的行没有调用默认构造函数或复制赋值运算符。

它从 construct 返回的临时对象复制构造 t3。

Answer 2

Foo t3 = Foo::construct(true); //It works! Why?

因为这不是默认初始化后赋值而是copy initialization

1) when a named variable (automatic, static, or thread-local) of a non-reference type T is declared with the initializer consisting of an equals sign followed by an expression.

所以根据这个说法：

If T is a class type and the cv-unqualified version of the type of other is T or a class derived from T, the non-explicit constructors of T are examined and the best match is selected by overload resolution. The constructor is then called to initialize the object.

还有这个：

If other is an rvalue expression, move constructor will be selected by overload resolution and called during copy-initialization. There is no such term as move-initialization.

在您的案例中使用了隐式声明的移动构造函数。

Answer 3

静态构造方法属于class，可以访问私有构造函数——这在实现工厂模式时经常使用。

静态方法class实例化

Static method class instantiation

c++

static-methods