On Tue, 09 Oct 2007 08:41:14 +0200, Marco Costalba <mcostalba@gmail.com> wrote:
I've tried to understand why it seems strange to me the proposed operator=() for assigning functions to overload:
void foo1(); int foo2(int);
overload f; f = &foo1; f = &foo2;
I came to the conclusion that all boils down to transitive property of equality. In other words until childhood we are teached that
If a == b and b == c, then a == c
Because the following line
if (a = b) // here = instead of == is intended assert(a==b);
should never fail for any properly defined operator=() and operator==() it derives that operator=() as proposed for our overload does not satisfy the above very intuitive concepts (because &foo1 != &foo2) so I would say it cannot be called 'idiomatic' for this case.
Marco
I consider operator=() misleading, too. I guess that for &foo1 != &foo2 you mean that their types are not convertible between each other. Marco -- Using Opera's revolutionary e-mail client: http://www.opera.com/mail/