I'm facing a problem in C++ :
#include <iostream>
class A
{
protected:
void some_func(const unsigned int& param1)
{
std::cout << "A::some_func(" << param1 << ")" << std::endl;
}
public:
virtual ~A() {}
virtual void some_func(const unsigned int& param1, const char*)
{
some_func(param1);
}
};
class B : public A
{
public:
virtual ~B() {}
virtual void some_func(const unsigned int& param1, const char*)
{
some_func(param1);
}
};
int main(int, char**)
{
A* t = new B();
t->some_func(21, "some char*");
return 0;
}
I'm using g++ 4.0.1 and the compilation error :
$ g++ -W -Wall -Werror test.cc
test.cc: In member function ‘virtual void B::some_func(const unsigned int&, const char*)’:
test.cc:24: error: no matching function for call to ‘B::some_func(const unsigned int&)’
test.cc:22: note: candidates are: virtual void B::some_func(const unsigned int&, const char*)
Why do I must specify that the call of some_func(param1) in class B is A::some_func(param1) ? Is it a g++ bug or a random message from g++ to prevent special cases I don't see ?
The problem is that in the derived class you are hiding the protected method in the base class. You can do a couple of things, either you fully qualify the protected method in the derived object or else you bring that method into scope with a using directive:
class B : public A
{
protected:
using A::some_func; // bring A::some_func overloads into B
public:
virtual ~B() {}
virtual void some_func(const unsigned int& param1, const char*)
{
A::some_func(param1); // or fully qualify the call
}
};
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