Legal to override same function in different bases?

[Old Wolf]

Is the following code well-defined, and outputs foo twice?

#include <iostream>

struct Base1 { virtual void foo() = 0; };
struct Base2 { virtual void foo() = 0; };

struct Derived: Base1, Base2
{
void foo() { std::cout << "foo" << std::endl; }
};

int main()
{
Derived d;
Base1 *b1 = &d;
Base2 *b2 = &d;
b1->foo(); b2->foo();
}

[gi4nospam]

On Oct 8, 9:52 am, Old Wolf <[e-mail]:

Is the following code well-defined, and outputs foo twice?

#include <iostream>

struct Base1 { virtual void foo() = 0; };
struct Base2 { virtual void foo() = 0; };

struct Derived: Base1, Base2
{
void foo() { std::cout << "foo" << std::endl; }
};

int main()
{
Derived d;
Base1 *b1 = &d;
Base2 *b2 = &d;
b1->foo(); b2->foo();
}

All compilers I have used will do exactly as you describe (output "foo
\nfoo\n"). It makes little sense to do otherwise doesn't it ?
Sometimes, however, I would like to prevent any override.

[James Kanze]

On Oct 8, 1:52 am, Old Wolf <[e-mail]:

Is the following code well-defined, and outputs foo twice?

#include <iostream>

struct Base1 { virtual void foo() = 0; };
struct Base2 { virtual void foo() = 0; };

struct Derived: Base1, Base2
{
void foo() { std::cout << "foo" << std::endl; }
};

int main()
{
Derived d;
Base1 *b1 = &d;
Base2 *b2 = &d;
b1->foo(); b2->foo();
}

Yes. It's fully defined. IIRC, this is even discussed some in
Stroustrup (one of the editions, anyway). The issue, of course,
is if Base1 is actuall GraphicObject, Base2 CardDeck, and the
function "draw()".

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James Kanze (GABI Software) email:ja*********@gmail.com
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