c++ - What does it mean to declare a variable as a function? -

in example below do

myclass (); 

i've been told function returns myclass neither of following lines work.

myclass b = a(); a.a = 1; 

so , can it?

#include "stdafx.h" #include "iostream" using namespace std;  class myclass {     public:         int a; };  int _tmain(int argc, _tchar* argv[]) {     myclass ();     // function? mean? can a?      int exit; cin >> exit;     return 0; } 

i've been told function returns myclass [...]

that function declaration. declares function called a , makes compiler aware of existence , signature (in case, function takes no arguments , returns object of type myclass). means may provide definition of function later on:

#include "iostream"  using namespace std;  class myclass {     public:         int a; };  int _tmain() {     myclass (); // *declares* function called "a" takes no                   // arguments , returns object of type myclass...      myclass b = a(); // right, long definition                      // function a() provided. otherwise, linker                      // issue undefined reference error.      int exit; cin >> exit;     return 0; }  myclass a() // ...and here definition of function {     return myclass(); }