#include <iostream>
namespace B {
struct A {
void foo() const { std::cout << "foo" << std::endl; }
};
void bar(const A &a) { a.foo(); }
void baz() { std::cout << "baz" << std::endl; }
}
int main(int argc,char **argv) {
B::A a;
bar(a);
B::baz(); // simply calling baz() does not work
return 0;
}
如果没有额外的资格,可以如何调用bar(a)？我原以为只有B :: bar(a)会编译.
当函数在命名空间内没有参数时,不会发生这种情况.
解决方法
When the postfix-expression in a function call is an unqualified-id,other namespaces not considered during the usual unqualified lookup may be searched,and in those namespaces,namespace-scope friend function or function template declarations not otherwise visible may be found. These modifications to the search depend on the types of the arguments (and for template template arguments,the namespace of the template argument).
和以下：
For each argument type T in the function call,there is a set of zero or more associated namespaces and a set of zero or more associated classes to be considered. The sets of namespaces and classes is determined entirely by the types of the function arguments..
If
Tis a class type (including unions),its associated classes are: the class itself; the class of which it is a member,if any; and its direct and indirect base classes. Its associated namespaces are the innermost enclosing namespaces of its associated classes.
实际上你甚至可以通过附上函数名来防止这种情况发生：
(bar)(a); // doens't compile
而
(B::bar)(a); // does compile
还要注意,这仅适用于最里面的命名空间,这意味着在以下情况下,您需要限定命名空间：
namespace B {
namespace C {
struct A {};
}
void bar(const C::A& a) { ... }
}
