CODEWITHSUNDEEP
c++oopinheritanceusing
Julien Vaslet
Julien Vaslet

Reputation: 1804

Why should I use the "using" keyword to access my base class method?

I wrote the below code in order to explain my issue. If I comment the line 11 (with the keyword "using"), the compiler does not compile the file and displays this error: invalid conversion from 'char' to 'const char*'. It seems to not see the method void action(char) of the Parent class in the Son class.

Why the compiler behave this way? Or have I done something wrong?

class Parent
{
    public:
        virtual void action( const char how ){ this->action( &how ); }
        virtual void action( const char * how ) = 0;
};

class Son : public Parent
{
    public:
        using Parent::action; // Why should i write this line?
        void action( const char * how ){ printf( "Action: %c\n", *how ); }
};

int main( int argc, char** argv )
{
    Son s = Son();
    s.action( 'a' );
    return 0;
}

Upvotes: 80

Views: 53968

Answers (5)

Scriabin
Scriabin

Reputation: 21

By using using, the names declared in base class are introduced into the namespace of derived class.

Then, when you declare the set of functions in the derived class, they are distinguished from those with identical parameter types in its base class by compilers by the type of the implicit this pointer.

During overload resolution, an argument that needs a class-type conversion, which will happen when a pointer to derived class is converted to a pointer to a base class, is of the lowest priority.

So the two functions with seemingly identical parameter list can be distinguished, and when calling them from an object of the derived type, the one locally declared will be a better (if not exact) match.

I am a newbie and please point out my misunderstanding.

Upvotes: 0

Bhupesh Pant
Bhupesh Pant

Reputation: 4349

If in a derived class any over loaded function is redefined then all the overloaded function in the base class is hidden. One way to include both the functionality is to avoid function overloading in classes. or You can use using keyword, as used.

Upvotes: 6

Amnon
Amnon

Reputation: 7772

Surprisingly this is standard behavior. If a derived class declares a method with the same name as a method defined by the base class, the derived class' method hides the base class' one.

See C++ FAQ

Upvotes: 21

sth
sth

Reputation: 229663

The action declared in the derived class hides the action declared in the base class. If you use action on a Son object the compiler will search in the methods declared in Son, find one called action, and use that. It won't go on to search in the base class's methods, since it already found a matching name.

Then that method doesn't match the parameters of the call and you get an error.

See also the C++ FAQ for more explanations on this topic.

Upvotes: 66

Dale Wilson
Dale Wilson

Reputation: 9434

A note of caution: The need to use a "using" in this situation is a red flag that your code may be confusing for other developers (after all it confused the compiler!). It is likely that you should rename one of the two methods to make the distinction clear to other programmers.

One possibility:

void action( const char how )
{ 
  takeAction( &how ); 
}
void action( const char * how )
{
  takeAction(how);
}
virtual void takeAction(const char * how) = 0;

Upvotes: 6

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