On Mon, 13 Jan 2003 07:48:24 +0100, Terje Slettebų <[EMAIL PROTECTED]> wrote:
>No, I don't think so. The previous sentence says: "The parameter list (void) >is equivalent to the empty parameter list." Then it follows with what is the >rule, except for this special case. The problem is the expression "except for this special case". The above is not a special case of parameter with void type. > IOW, it's not a parameter type in the >special case, and it's not otherwise, either. Exactly. >How would you propose to clarify it? Since you asked, I would have added the special case to the grammar, where one can unambiguously specify whether void is intended literally or not. Example: parameter-declaration-clause: parameter-declaration-listopt ...opt parameter-declaration-list , ... empty-parameter-list-specifier empty-parameter-list-specifier: void Then, since I care clarity (:-)) I would have written: A parameter-declaration-clause consisting of the empty-parameter-list-specifier is equivalent to the empty parameter list. [Note: by grammar, the empty-parameter-list-specifier can only be (at translation phase 7) the keyword "void" and not e.g. a type-id of type void] [...] >> template <typename T1, typename T2, typename T3> >> int f(T1 t1, T2 t2, T3 t3); >> >> >>becomes a generator of variadic functions with a fixed maximum number >>of arguments. > >Really? How would you impleent that? Yup. Forget it, this was just a result of somnolence. In practice, while sleeping at the keyboard, it occurred to me the wonderful idea that if you allow f (T); with T=void than you have a variadic function with zero or one arguments. Hence the "generalization"... Genny. _______________________________________________ Unsubscribe & other changes: http://lists.boost.org/mailman/listinfo.cgi/boost
