我有以下代码:
struct M { friend void f() {} M() { f(); // error: 'f' was not declared in this scope } }; int main() { M m; }
g 4.8和clang3.4都无法编译,因为f在M中是不可见的,所以他们说.
然而,该标准给出了类似代码的示例
class M { friend void f() { } // definition of global f,a friend of M,// not the definition of a member function };
并说
A
friend
function defined in a class is in the (lexical) scope of the
class in which it is defined.
(ISO / IEC 14882:2011 11.3朋友[class.friend] p6,p7)
从这里我无法理解编译器如何找不到f,它被定义在同一个类中.
两个编译器都有同样的错误.
那么,我错过了什么?
解决方法
朋友声明指出,在周围的命名空间中,一个名为f的函数是该类的朋友;但是它不会将名字f引入命名空间.它不可用(除了通过参数相关的查找),直到它在命名空间中声明.
相关规则为C 11 7.3.1.2/3:
If a
friend
declaration in a non-local class first declares a class or function the friend class or function is a member of the innermost enclosing namespace. The name of the friend is not found by unqualified lookup or by qualified lookup until a matching declaration is provided in that namespace scope.