c++ - Compiler can't find base class method when called from derived, and the derived defines same named method with additional parameter -

here's link ideone simple code paste: http://ideone.com/bbck3b .

the base class has paramtereless function, whereas derived has 1 parameter. public.

why compiler fails find a::foo() when called instance of b?

the code:

#include <iostream> using namespace std;  class { public:     virtual void foo()     {         cout << "a::foo" << endl;     } };  class b : public { public:     void foo(int param)     {         cout << "b::foo " << param << endl;     } };  int main() {     b b;     b.foo(); } 

the compiler error:

prog.cpp: in function ‘int main()’: prog.cpp:25:11: error: no matching function call ‘b::foo()’      b.foo();            ^ prog.cpp:25:11: note: candidate is: prog.cpp:16:10: note: void b::foo(int)      void foo(int param)           ^ prog.cpp:16:10: note:   candidate expects 1 argument, 0 provided 

this standard c++ behaviour: base class method hidden derived-class method of same name, regardless of arguments , qualifiers. if want avoid this, have explicitly make base-class method(s) available:

class b : public { public:   void foo(int param)  // hides a::foo()   {     cout << "b::foo " << param << endl;   }   using a::foo;        // make a::foo() visible again };