[boost] Re: dangerous_cast<>

From: Gennaro Prota (gennaro_prota_at_[hidden])
Date: 2002-12-07 11:16:42

On 07 Dec 2002 12:26:35 +0100, Gabriel Dos Reis
<gdr_at_[hidden]> wrote:

>5.2.10/1:
>  The result of the expression reinterpret_cast<T>(v) is the result of
>  converting the expression v to type T. [...]

Are you saying you really understand what that phrase means? Can you
explain it, and say please what notion do you assume as "primitive"
(e.g. "converting an expression)?

>and converting a Foo* to void* is well-defined.

I don't think you can say whether a conversion is well-defined or not
in C++ without considering *how* it is done.

Maybe these are minor points in the context of your reasoning. Sorry
if that is the case, but I tend to stop reading long posts when I hit
(what I think is) the first major problem.

Genny.