c – 为什么移动构造函数涉及到此处

我有这段C代码：

class Args {};

class MyClass {
  public:
  MyClass(Args& a) {}
  MyClass(MyClass &&) = delete;
};

int main() {

  Args a;
  MyClass c1 = MyClass(a);
  MyClass c2 = a;
  MyClass c3(a);

  return 0;
}

这不会编译,因为对象c1和c2的构造似乎涉及类的移动构造函数：

错误：使用已删除的函数’MyClass :: MyClass(MyClass&&)’

似乎编译器想要创建临时对象,然后将它们移动到c1和c2.为什么会这样？不应该所有三个语句都只调用MyClass(Args& a)构造函数吗？

另一方面,如果我创建移动构造函数,程序编译正常,移动构造函数永远不会被调用！

解决方法

见 copy elision：

Under the following circumstances,the compilers are permitted,but not required to omit the copy- and move- (since C++11) construction of class objects even if the copy/move (since C++11) constructor and the destructor have observable side-effects. This is an optimization: even when it takes place and the copy-/move-constructor is not called,it still must be present and accessible (as if no optimization happened at all),otherwise the program is ill-formed.

从C 17开始：

They need not be present or accessible,as the language rules ensure that no copy/move operation takes place,even conceptually.

原文地址：https://www.jb51.cc/c/112969.html