257

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.

6
  • 1
    Duplicate: stackoverflow.com/questions/411103/…
    – psychotik
    Commented Oct 27, 2009 at 4:30
  • 10
    brilliant question, i only discovered this recently too Commented Oct 27, 2009 at 7:30
  • 12
    I think Bjarne (from the link Mac posted) put it best in one sentence: "In C++, there is no overloading across scopes - derived class scopes are not an exception to this general rule."
    – sivabudh
    Commented Feb 11, 2010 at 17:56
  • 7
    @Ashish That link is broken. Here's the correct one (as of now) - stroustrup.com/bs_faq2.html#overloadderived
    – nsane
    Commented Oct 25, 2015 at 11:37
  • 3
    Also, wanted to point out that obj.Base::gogo(7); still works by calling the hidden function. Commented Jan 30, 2018 at 21:19

4 Answers 4

445

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 counter-intuitive, 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 : public Base{
    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 relationship 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.

18
  • 5
    Great answer! Also, as a practical matter, compilation would probably get a lot slower if the name search had to go all the way to the top every time.
    – Drew Hall
    Commented Oct 27, 2009 at 11:24
  • 6
    (Old answer, I know.) Now will nullptr I would object to your example by saying "if you wanted to call the void* version, you should use a pointer type". Is there a different example where this can be bad?
    – GManNickG
    Commented May 17, 2011 at 19:26
  • 2
    @DrewHall: That seems like a data structure problem. The compiler should be smart enough to store the base and derived methods of a derived class in a hash table or similar structure for quick lookup instead of crawling the hierarchy. Commented Jun 6, 2012 at 20:52
  • 3
    The name hiding isn't really evil. The "is-a" relationship is still there, and available through the base interface. So maybe d->foo() won't get you the "Is-a Base", but static_cast<Base*>(d)->foo() will, including dynamic dispatch.
    – Kerrek SB
    Commented Jan 9, 2014 at 13:36
  • 16
    This answer is unhelpful because the example given behaves the same with or without hiding: D::foo(int) will be called either because it is a better match or because it has hidden B:foo(int). Commented Jul 8, 2014 at 3:06
55

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.

1
  • 11
    OP: "Why does an overridden function in the derived class hide other overloads of the base class?" This answer: "Because it does". Commented Jul 8, 2014 at 3:58
13

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.

0
2

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/

0

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