c++ - Weird pointer to member function syntax -

i understand how declare type of function:

typedef void (typedef_void_f)(); // typedef_void_f void() using alias_void_f     = void(); // alias_void_f void() 

and can used declare function pointers:

void function() { std::cout << __pretty_function__ << '\n'; }  typedef_void_f *a = function; // pointer void() alias_void_f   *b = function; // pointer void() 

for member function pointers syntax more complicated:

struct s { void function() { std::cout << __pretty_function__ << '\n'; } };  typedef void (s::*typedef_void_m_f)(); using  alias_void_m_f = void (s::*)();  typedef_void_m_f c = &s::function; // pointer s void() member function alias_void_m_f   d = &s::function; // pointer s void() member function 

this understanding of function pointers in c++ , thought enough.

but in p0172r0 technical paper i've found syntax i'm not familiar:

struct host {    int function() const; };  template <typename type> constexpr bool test(type host::*) { // <---- this??     return is_same_v<type, int() const>; }  constexpr auto member = &host::function;  test(member); 

as understand code, test function splits type of function type of object function belongs, in template test function type template parameter void() if try following:

void my_test(void() s::*) {}  my_test(&s::function); 

i bunch of syntax errors:

error: variable or field 'my_test' declared void  void my_test(void() s::*) {}                    ^ error: expected ')' before 's'  void my_test(void() s::*) {}                      ^ error: 'my_test' not declared in scope      my_test(&s::function); 

so obvious i'm not understanding p0172r0's test function syntax.

can explain details of template <typename type> constexpr bool test(type host::*) syntax?

type host::* pointer class data member. type type of class member , host::* means pointer member of host. type host::* accepts pointer of member of host