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

c++, compilation, overloading

Solution

This is standard C++ behaviour: the base class method is hidden by a derived-class method of the same name, regardless of the arguments and qualifiers. If you want to avoid this, you have to explicitly make the base-class method(s) available:

class B : public A
{
  public:
    void foo(int param)  // hides A::foo()
    {
        cout << "B::foo " << param << endl;
    }
    using A::foo;        // makes A::foo() visible again
};

Problem

Here's a link to ideone with a simple code paste: http://ideone.com/BBcK3B . The base class has a paramtereless function, whereas the derived has one with a parameter. Everything is public. Why the compiler fails to find A::foo() when called from instance of B? The code: ``` #include <iostream> using namespace std; class A { public: virtual void foo() { cout << "A::foo" << endl; } }; class B : public A { 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 for call to ‘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 ```

Original source

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