C++ compile time polymorphism doubt ?

Question

Below program contains two show() functions in parent and child classes, but first show() function takes FLOAT argument and second show() function takes INT argument.

.If I call show(10.1234) function by passing float argument, it should call class A's show(float a) function , but it calls class B's show(int b).

#include<iostream>
using namespace std;

class A{
        float a;
public:
        void show(float a)
        {
                this->a = a;
                cout<<"\n A's show() function called : "<<this->a<<endl;
        }
};

class B : public A{
        int b;
public:
        void show(int b)
        {
                this->b = b;
                cout<<"\n B's show() function called : "<<this->b<<endl;
        }
};

int main()
{
        float i=10.1234;
        B Bobject;
        Bobject.show((float) i);
        return 0;
}

Output:

B's show() function called : 10

Expected output:

A's show() function called : 10.1234

Why g++ compiler chosen wrong show() function i.e class B's show(int b) function ?

Answer 1

There is no polymorphism involved here. You should declare your functions virtual to make them polymorphic, and make them have the same signature.

Answer 2

If you have a function in a derived class that has the same name as a function in the base class, it hides all of the functions in the base class. You either need to rename your function, or use a using declaration in your derived class:

using A::show;

or, you can explicitly call the base class function:

Bobject.A::show(i);

Answer 3

Use a using declaration.

class B : public A{
        int b;
public:
        using A::show;
        …
};

Answer 4

You're mixing things:
If you had 2 functions in class A with the following signatures:
void Show(int a);
void Show(float a);

Then the compiles would have chosen the "correct" function.

Answer 5

When you define a name in a derived class, it hides the name from the base class.

You can achieve what you want by adding this to B's definition:

using A::show;

That will let you use both A's show() and B's show().