C++ base template class virtual methods doesn't appear in derived? -

this question has answer here:

• overriding base's overloaded function in c++ [duplicate] 3 answers
• function same name different signature in derived class 2 answers

lest consider following code:

#include <iostream>  template <typename t_val>     // not used template argument struct foo {        int x;    };  template <typename t_val> struct bar {        int x;    };  template <typename t_value> struct {     // 2 overloaded data() members:      virtual void data(foo<t_value>) {         std::cout << "foo empty\n";     }      virtual void data(bar<t_value>) {         std::cout << "bar empty\n";     } };   template <typename t_value> struct b : public a<t_value> {     void data(foo<t_value>) {         std::cout << "smart foo impl\n";     } };   int main(void) {      b<float> theobject;     bar<float> bar;      theobject.data(bar); // expect virtual base call here     return 0; } 

the compiler message is:

tmpl.cpp: in function 'int main()': tmpl.cpp:43:14: error: no matching function call 'b<float>::data(bar<float>&)'   theobject.data(bar);                     ^ 

void data(bar<t_value>) -> void data(bar<t_value>&) not change anything

why derived class b has no virtual void data(bar&)?

the other data hidden.

do

template <typename t_value> struct b : public a<t_value> {     using a<t_value>::data; // add line     void data(foo<t_value>) {         std::cout << "smart foo impl\n";     } };