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

Question

Consider the code :

#include 

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

    v         


        

Answer 1

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/

Answer 2

Judging by the wording of your question (you used the word "hide"), you already know what is going on here. The phenomenon is called "name hiding". For some reason, every time someone asks a question about why name hiding happens, people who respond either say that this called "name hiding" and explain how it works (which you probably already know), or explain how to override it (which you never asked about), but nobody seems to care to address the actual "why" question.

The decision, the rationale behind the name hiding, i.e. why it actually was designed into C++, is to avoid certain counterintuitive, unforeseen and potentially dangerous behavior that might take place if the inherited set of overloaded functions were allowed to mix with the current set of overloads in the given class. You probably know that in C++ overload resolution works by choosing the best function from the set of candidates. This is done by matching the types of arguments to the types of parameters. The matching rules could be complicated at times, and often lead to results that might be perceived as illogical by an unprepared user. Adding new functions to a set of previously existing ones might result in a rather drastic shift in overload resolution results.

For example, let's say the base class B has a member function foo that takes a parameter of type void *, and all calls to foo(NULL) are resolved to B::foo(void *). Let's say there's no name hiding and this B::foo(void *) is visible in many different classes descending from B. However, let's say in some [indirect, remote] descendant D of class B a function foo(int) is defined. Now, without name hiding D has both foo(void *) and foo(int) visible and participating in overload resolution. Which function will the calls to foo(NULL) resolve to, if made through an object of type D? They will resolve to D::foo(int), since int is a better match for integral zero (i.e. NULL) than any pointer type. So, throughout the hierarchy calls to foo(NULL) resolve to one function, while in D (and under) they suddenly resolve to another.

Another example is given in The Design and Evolution of C++, page 77:

class Base {
    int x;
public:
    virtual void copy(Base* p) { x = p-> x; }
};

class Derived{
    int xx;
public:
    virtual void copy(Derived* p) { xx = p->xx; Base::copy(p); }
};

void f(Base a, Derived b)
{
    a.copy(&b); // ok: copy Base part of b
    b.copy(&a); // error: copy(Base*) is hidden by copy(Derived*)
}

Without this rule, b's state would be partially updated, leading to slicing.

This behavior was deemed undesirable when the language was designed. As a better approach, it was decided to follow the "name hiding" specification, meaning that each class starts with a "clean sheet" with respect to each method name it declares. In order to override this behavior, an explicit action is required from the user: originally a redeclaration of inherited method(s) (currently deprecated), now an explicit use of using-declaration.

As you correctly observed in your original post (I'm referring to the "Not polymorphic" remark), this behavior might be seen as a violation of IS-A relationsip between the classes. This is true, but apparently back then it was decided that in the end name hiding would prove to be a lesser evil.

Answer 3

This is "By Design". In C++ overload resolution for this type of method works like the following.

• Starting at the type of the reference and then going to the base type, find the first type which has a method named "gogo"
• Considering only methods named "gogo" on that type find a matching overload

Since Derived does not have a matching function named "gogo", overload resolution fails.

Answer 4

The name resolution rules say that name lookup stops in the first scope in which a matching name is found. At that point, the overload resolution rules kick in to find the best match of available functions.

In this case, gogo(int*) is found (alone) in the Derived class scope, and as there's no standard conversion from int to int*, the lookup fails.

The solution is to bring the Base declarations in via a using declaration in the Derived class:

using Base::gogo;

...would allow the name lookup rules to find all candidates and thus the overload resolution would proceed as you expected.