Strange behaviour when calling virtual functions

Question

I don't understand what is wrong with this code. It looks like an incredible trap !

This code :

class Foo
{
  public:
      virtual double foo(double x) const = 0;
              double foo(int x) const { return (double)(x + x); }
};

class Bar : public Foo
{
    public:
        virtual double foo(double x) const { return x * x; }
};

int main()
{
    Bar* b = new Bar;
    Foo* f = b;
    std::cout << b->foo(3) << " " << f->foo(3) << std::endl;
    std::cout << b->foo(5.0) << " " << f->foo(5.0) << std::endl;
    return 0;
}

prints the following output :

9 6
25 25

I deduce that Bar::foo(double) const is called with an implicit cast when the type of the pointer is Bar*. But why such a thing is possible without any warning ?

I work with GCC 4.7.2. I compiled with g++ -Wall foobar.cpp -o foobar.exe

I think it would be better if you choose your second number another than 2. Because 2*2=4 and 2+2=4. — nabroyan, Jun 10 '15 at 08:55
I'm curious, do you have the same issue if the return value of `foo(int)` is also `int`? @nabroyan indeed, it would be even better to use the same value (but different type): `3` and `3.0`. — Kryptos, Jun 10 '15 at 08:56
Possible duplicate of http://stackoverflow.com/questions/1628768/why-does-an-overridden-function-in-the-derived-class-hide-other-overloads-of-the and http://stackoverflow.com/questions/411103/function-with-same-name-but-different-signature-in-derived-class — Kryptos, Jun 10 '15 at 09:01

score 9 · Accepted Answer · answered Jun 10 '15 at 08:56

This is due to name hiding.

When you declare a function called foo in Bar, you hide all declarations with the same name in Foo.

As such, when the static type of the pointer is Bar, the compiler only finds the version in Bar which takes a double, so it implicitly converts the int to satisfy this.

If you want the int version in Foo to be visible, add a using declaration:

class Bar : public Foo
{
    public:
        using Foo::foo;
//      ^^ makes the versions in Foo visible
        virtual double foo(double x) const { return x * x; }
};

score 2 · Answer 2 · answered Jun 10 '15 at 08:55

When the type is Bar*, only one version of the method is visible, the one with the double parameter.

Base methods with the same name (but different signature) are hidden.

To make them available, you can use using Foo::foo in the derived class.

Depending on your compiler, I think you might also get warnings regarding the implicit conversion, or the fact that you apparently want to call a hidden method.

score 1 · Answer 3 · answered Jun 10 '15 at 08:56

In Foo there are two overloads of foo, one that takes a double and another that takes an int.

In Bar there is one overload of foo, the one that takes a double. This overload hides all functions with the same name from the base classes. This is called name hiding.

A fix would be to employ using declaration to bring the other foo overloads from the base class in scope of Foo derived class:

class Bar : public Foo
{
    public:
        using Foo::foo; 
        virtual double foo(double x) const { return x * x; }
};

Strange behaviour when calling virtual functions

3 Answers3

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