Why does an overridden function in the derived class hide other overloads of the base class?

user_74214 0
Viewed 50398

Consider the code : 

#include <stdio.h>

class Base {
public: 
    virtual void gogo(int a){
        printf(" Base :: gogo (int) \n");
    };

    virtual void gogo(int* a){
        printf(" Base :: gogo (int*) \n");
    };
};

class Derived : public Base{
public:
    virtual void gogo(int* a){
        printf(" Derived :: gogo (int*) \n");
    };
};

int main(){
    Derived obj;
    obj.gogo(7);
}

Got this error :

>g++ -pedantic -Os test.cpp -o test
test.cpp: In function `int main()':
test.cpp:31: error: no matching function for call to `Derived::gogo(int)'
test.cpp:21: note: candidates are: virtual void Derived::gogo(int*) 
test.cpp:33:2: warning: no newline at end of file
>Exit code: 1

Here, the Derived class's function is eclipsing all functions of same name (not signature) in the base class. Somehow, this behaviour of C++ does not look OK. Not polymorphic.

overriding polymorphism c++
4 Answers

Name hiding makes sense because it prevents ambiguities in name resolution.

Consider this code:

class Base
{
public:
    void func (float x) { ... }
}

class Derived: public Base
{
public:
    void func (double x) { ... }
}

Derived dobj;

If Base::func(float) was not hidden by Derived::func(double) in Derived, we would call the base class function when calling dobj.func(0.f), even though a float can be promoted to a double.

Reference: http://bastian.rieck.ru/blog/posts/2016/name_hiding_cxx/

user_7200940
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