c++ - 已删除的构造函数/运算符的可访问性是否重要？

Question

如果一个类型的默认成员被删除，声明的可访问性是否会有所不同？

class FooA {
public:
  FooA() = delete;
  FooA(FooA const&) = delete;
  FooA& operator=(FooA const&) = delete;
}

class FooB {
private:
  FooB() = delete;
  FooB(FooB const&) = delete;
  FooB& operator=(FooB const&) = delete;
}

class FooC {
protected:
  FooC() = delete;
  FooC(FooC const&) = delete;
  FooC& operator=(FooC const&) = delete;
}

Answer 1

虽然accessibility and deletedness are orthogonal ，很难看出在您提出的情况下会有什么实际差异。