Nope, did now. On 8/8/04 11:30 AM, "David Abrahams" <dave@boost-consulting.com> wrote:
Did you send this to the OP ("Dave" <better_cs_now@yahoo.com>)?
Jeremy Graham Siek <jsiek@osl.iu.edu> writes:
Hello Dave,
Sorry about the confusion. I should have distinguished between how this works in C++, and what is theoretically possible. The contravariance rule I spoke of describes what makes sense theoretically. However, C++ is more conservative in this case and restricts parameter types of overloads to be the same type. This is a similar issue to covariant return types. Early on, C++ did not allow them, even though it made sense theoretically, then later on covariant return types were allowed.
Best Regards, Jeremy
On 8/8/04 2:09 AM, "Dave" <better_cs_now@yahoo.com> wrote:
Hello all,
Quoting from page 24 of "The Boost Graph Library; User Guide and Reference Manual":
"It turns out that by the contravariance subtyping rule, the parameter type in the derived classes member function must be either the same type or a base class of the type as the parameter in the base class."
Now please consider this code:
#include <iostream>
struct base_1 {}; struct derived_1: base_1 {};
struct base_2 { virtual void foo(derived_1 *p) {std::cout << "base_2::foo()\n";} };
struct derived_2: base_2 { virtual void foo(base_1 *p) {std::cout << "derived_2::foo()\n";} };
int main() { base_2 *ptr = new derived_2; ptr->foo(new derived_1); }
This outputs base_2::foo(). Why?
The quoted passage implies derived_2::foo() should override base_2::foo(). Clearly, it does not. Furthermore, I cannot find anything in the Standard that indicates it should. So, I am clearly misinterpreting the quoted passage. Can anybody explain to me what was meant in that passage?
Thank you, Dave
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