c++ - 为什么函数名不能与返回名类型相同？

Question

不确定我缺少什么，但是为什么不能编译？

class Game
{
};

class Actor
{
   Game* mGame;
   Game* Game() { return mGame; }
};

int main()
{
  Actor a();

  return 0;
}

g++ -std=c++17 main.cpp -o 测试

main.cpp:8:10: error: declaration of ‘Game* Actor::Game()’ changes meaning of ‘Game’ [-fpermissive]
    8 |    Game* Game() { return mGame; }
      |          ^~~~
main.cpp:1:7: note: ‘Game’ declared here as ‘class Game’
    1 | class Game
      |      

显然，如果我将函数更改为 GetGame，则没有问题。只是想知道为什么 - 我错过了什么？

谢谢!

Answer 1

由于添加了语言律师标签，并且对于这是否由标准定义似乎存在一些混淆，以下是标准对现有答案的补充说明。

在 [basic.scope.hiding] 中:

If a class name (11.2) or enumeration name (9.6) and a variable, data member, function, or enumerator are declared in the same declarative region (in any order) with the same name (excluding declarations made visible via using-directives (6.4.1)), the class or enumeration name is hidden wherever the variable, data member, function, or enumerator name is visible.

此外，在 [class.name] 中，我们发现:

If a class name is declared in a scope where a variable, function, or enumerator of the same name is also declared, then when both declarations are in scope, the class can be referred to only using an elaborated-type-specifier (6.4.4). [Example:

struct stat {
  // ...
};

stat gstat;                // use plain stat to define variable

int stat(struct stat*);    // redeclare stat as function

void f() {
  struct stat* ps;         // struct prefix needed to name struct stat
  stat(ps);                // call stat()
}

— end example]

所以，在 f的范围内, stat指的是该名称的函数，而 struct stat (详细的类型说明符)指的是类。

或者在 OP 的示例中，在 Actor 的范围内, Game指的是该名称的(成员)函数，而 class Game (详细的类型说明符)指的是类。

请注意，或者， ::Game可以用来引用类。

最后，在 [class.name] 的更深处，有一个关于编写此类代码的相关引用:

4 [Note: The declaration of a class name takes effect immediately after the identifier is seen in the class definition or elaborated-type-specifier. For example, class A * A; first specifies A to be the name of a class and then redefines it as the name of a pointer to an object of that class. This means that the elaborated form class A must be used to refer to the class. Such artistry with names can be confusing and is best avoided. — end note]

因此，该标准不仅完全涵盖了这种情况 - 它还建议不要编写此类令人困惑的代码。