我有以下代码：
//mystd plays a role of the std namespace
//which does not allow any new overloads in it
//so we define one in the global namespace
namespace mystd {
template<typename T>
struct A {};
}
//if one uncomment this, everything works as designed
//int f(const mystd::A<int>& v);
//this some namespace which is supposed to be the main one
namespace a {
template<typename T>
int f(const T& v) {
return 0;
}
template<typename T>
int operator-(const T& v) {
return 0;
}
}
//this should be different from ::a
//so that `using namespace a;` would import global functions
//as well as all functions from namespace a, since
//the global namespace is the nearest common namespace
//see http://www.open-std.org/jtc1/sc22/open/n2356/dcl.html#namespace.udir
namespace b {
template<typename T>
int Call(const T& v) {
using namespace ::a;
return -v + f(v);
}
}
int operator-(const mystd::A<int>&) {
return 1;
}
int f(const mystd::A<int>&) {
return 2;
}
int main() {
//why this returns 1, not 3?
//what is the difference between operator-() & f()?
return b::Call(mystd::A<int>());
}
这个想法是模仿gtest运算符的行为＆lt;＆lt; ＆安培; PrintTo，因此可以重载某些std类的行为，这是不能通过ADL完成的，因为不允许向std命名空间添加任何内容。但是，如果我尝试使用通常的函数f()，那么f(mystd::A())的重载被认为定义太晚（如果有人评论gcc给出的f()的模板定义note: ‘int f(const mystd::A<int>&)’ declared here, later in the translation unit）
问题是为什么operator-()和f()的行为不同？