Cannot find baseclass function if baseclass has two functions of same name

Question

I have a baseclass that has two functions of same name but with different signatures in a 2-level inheritance.

struct A {
    virtual void f(int) { }
    virtual void f(int, int) { };
    virtual void f1(int) { }
};

struct B: public A { };

struct C: public B {
  void f(int, int) { }
  void f1(int) { }
};

int main() {
 C obj;
 obj.f1(0);
 obj.f(0,0);

 obj.f(0);    // (1) cannot be found
 obj.B::f(0); // (2) works

}

I would have expected my compiler (gcc-4.3.2) to find the correct definition at (1), but I get

g++     main.cpp   -o main
main.cpp: In function 'int main()':
main.cpp:20: error: no matching function for call to 'C::f(int)'
main.cpp:10: note: candidates are: virtual void C::f(int, int)
distcc[2200] ERROR: compile main.cpp on localhost failed
make: *** [main] Error 1

(2) on the other hand works.

What do I need to fix to make (1) work in general?

+1 for well-written post with complete example

John Dibling
– John Dibling

2011-01-19 23:13:35 +00:00
Commented Jan 19, 2011 at 23:13 — John Dibling
– John Dibling, Commented Jan 19, 2011 at 23:13

Lightness Races in Orbit · Accepted Answer · 2011-01-19 23:16:51Z

6

Write using A::f inside the definition of C.

You are a victim of name hiding! void C::f(int, int) hides void A::f(int), just because.

answered Jan 19, 2011 at 23:16

Lightness Races in Orbit

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Comments

Zan Lynx · Accepted Answer · 2011-01-19 23:18:33Z

5

The C++ name lookup rules have it so that if a name is redefined in one scope, all overloads of that name are hidden.

But you can use using to help. Like this:

class A {
    public:
    int f(int x) { cout << "A::f 1\n"; }
    int f(int x, int y) { cout << "A::f 2\n"; }
};

class B : public A {
    public:
    using A::f;
    int f(int x) { cout << "B::f 1\n"; }
};

int main()
{
    B b;

    b.f(27, 34);

    return 0;
}

Output is:

A::f 2

answered Jan 19, 2011 at 23:18

Zan Lynx

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Comments

John Dibling · Accepted Answer · 2011-01-19 23:16:38Z

3

The short answer to "why" is "because that's how overloading works." You are hiding the f(int) overload in C. The longer answer is much longer.

You can un-hide it by doing this:

struct C: public B {
  using A::f;
  void f(int, int) { }
  void f1(int) { }
};

answered Jan 19, 2011 at 23:16

John Dibling

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5 Comments

Benjamin Bannier Over a year ago

Can you tell us the longer answer?

John Dibling Over a year ago

@honk: Not right now, I'm short on time. But I'll enlist some help with this.

Fred Larson Over a year ago

It's not long, but here's an explanation: stackoverflow.com/questions/888235/…

Benjamin Bannier Over a year ago

Thanks @Fred Larson, I should have found that.

sbi Over a year ago

@John: Never let answering get in the way of beer-o-clock, right? :)

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