为什么我不能有 return 类型与 void* 协变？

Question

为什么下面的代码是协方差错误？ T * 不是与 void *... 协变的吗？

struct Base { virtual void *foo(); };
struct Derived : Base { int *foo(); };

海湾合作委员会说：

invalid covariant return type for 'virtual int* Derived::foo()'

Answer 1

[class.virtual]/p7，强调我的：

The return type of an overriding function shall be either identical to the return type of the overridden function or covariant with the classes of the functions. If a function D::f overrides a function B::f, the return types of the functions are covariant if they satisfy the following criteria:

both are pointers to classes, both are lvalue references to classes, or both are rvalue references to classes [footnote omitted]

[...]

来自 D&E 的第 294-5 页：

Afer some consideration of the alternatives, we decided to allow overriding of a B* by a D* and of a B& by a D& where B is an accessible base of D. In addition, const can be added or subtracted wherever that is safe. We decided not to relax the rules to allow technically feasible conversions such as a D to an accessible base B, a D to an X for which D has a conversion, int* to void*, double to int, etc. We felt that the benefits from allowing such conversions through overriding would not outweigh the implementation cost and the potential for confusing users.

Answer 2

协变 return 类型特征是派生的 class 为重写函数提供更多 specific/narrower return 类型。派生的 class return 类型被认为是协变的。

int* 不是 void* 类型，而像这样的东西确实描述了协变 return 类型：

struct Base {
  virtual void *foo();
  virtual Base* func();
};
struct Derived : Base {
//  int *foo();
  Derived* func(); //COVARIANT RETURN TYPE
};

Answer 3

void* 和 T* 之间的协方差是不允许的，因为：

1.缺乏一致性.

当前的协方差方式很容易理解，不会造成混淆。
试想 void* 类型的协方差是允许的。对于 1 阶段的推导很好，但它会造成混乱。例如：

struct void_ { virtual void* foo (); };
struct int_ : void_ { virtual int* foo (); };
struct double_ : int_ { virtual double* foo (); };  // int* to double* or void* to double*

在struct double_的第3层，用户会疑惑，虽然double*到int*都不可能，为什么代码还在编译呢？查了最上面的classvoid_才知道是因为double*到void*是"covariant"。编译器也一样:-)

2。引用问题

在 classes 的情况下，返回 B&/D*/DD* 是可能的。但是 void& 不可能发生同样的事情，因此 int& 等等

3。混合协方差

如果允许 void* 那么无意中也允许跟随。

struct void_ { virtual void* foo (); };
struct Base : void_ { virtual Base* foo (); };
struct Derived : Base { virtual int* foo (); }; // Look at `int*`

这会造成混乱。

为什么我不能有 return 类型与 void* 协变？

Why can't I have return types covariant with void*?

c++

return-type

covariance