Why is 'X x; x();' allowed, when 'X' defines a conversion to function pointer, but not, when it defines a conversion to a functor?

Question

void f(int){}
typedef void (*f_ptr)(int);

struct Functor{
  void operator()(int){}
};

struct X{
  operator f_ptr(){ return f; }
};

struct Y{
  operator Functor(){         


        

Answer 1

x(5); // works ?!

This implicitly casts x to an f_ptr and calls that. C++11 standard:

§ 13.3.1.1.2 Call to object of class type [over.call.object]

2) In addition, for each non-explicit conversion function declared in T of the form

operator conversion-type-id ( ) attribute-specifier-seqopt cv-qualifier ;

[…where conversion-type-id denotes the type “pointer to function of (P1,...,Pn) returning R”…]

---

y(5); // doesn't ?!

The standard doesn't mention anything about implicit conversion to class types that overload operator() (aka functors), which implies that the compiler doesn't allow that.

You must cast it explicitly:

static_cast<Functor>(y)(5);