N3797 :: 12.8 / 11 [class.copy]部分说：
An implicitly-declared copy/move constructor is an inline public
member of its class. A defaulted copy/ move constructor for a class X
is defined as deleted (8.4.3) if X has:[…]
— a non-static data
member of class type M (or array thereof) that cannot be copied/moved
because overload resolution (13.3),as applied to M’s corresponding
constructor,results in an ambiguity or a function that is deleted or
inaccessible from the defaulted constructor
关于相应的复制/移动构造函数的歧义的第一个案例非常清楚.我们可以写下面的内容：
#include <iostream>
using namespace std;
struct A
{
A(){ }
A(volatile A&){ }
A(const A&,int a = 6){ }
};
struct U
{
U(){ };
A a;
};
U u;
U t = u;
int main(){ }
反映这一点.但是,默认构造函数中删除或无法访问的函数或函数是什么？使用默认构造函数无法访问的函数有什么作用？你能举一个反映这个的例子吗？
解决方法
简单的说：
struct M { M(M const&) =delete; };
struct X { X(X const&) =default; M m; }; // X(X const&) is actually deleted!
隐式声明的函数也被视为“默认”([dcl.fct.def.default] / 5);一个更熟悉的前C 11例子可能是这样的：
struct M { protected: M(M const&); };
struct X { M m; }; // X's implicit copy constructor is deleted!
请注意,如果在声明函数后显式默认该函数,如果函数将被隐式删除([dcl.fct.def.default] / 5),则程序格式错误：
struct M { M(M const&) =delete; };
struct X { X(X const&); M m; };
X::X(X const&) =default; // Not allowed.
