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
This is due to name hiding.
When you declare a function called
foo
inBar
, you hide all declarations with the same name inFoo
.As such, when the static type of the pointer is
Bar
, the compiler only finds the version inBar
which takes adouble
, so it implicitly converts theint
to satisfy this.If you want the
int
version inFoo
to be visible, add ausing
declaration: