Re: writing MyClass::Operator unsigned long()

From:

"Howard" <alicebt@hotmail.com>

Newsgroups:

comp.lang.c++

Date:

Thu, 08 Jun 2006 17:29:27 GMT

Message-ID:

<X7Zhg.150370$Fs1.50511@bgtnsc05-news.ops.worldnet.att.net>

"Victor Bazarov" <v.Abazarov@comAcast.net> wrote in message
news:e69lm2$pep$1@news.datemas.de...

Howard wrote:

"Victor Bazarov" wrote:

   struct s {
       operator unsigned long() { return 42UL; }
   };

   int main() {
       unsigned long ul = s();
   }

[...correct analysis...]

Is my analysis (guess) as to its meaning correct? Does this syntax
have a name, so I could look it up in my book (The C++ Programming
Language; Stroustrup)?

Which one? :-)

In the statement

unsigned long ul = s();

the 's()' part is called "explicit type conversion (functional notation)",
and there is nothing between the parentheses (*). This expression yields
a *value-initialised* rvalue (a temporary) of type 's'.

That's the info was looking for.

(*) The parentheses can contain a comma-delimited expression list, or
   a single expression. The expression[s] become arguments to the
   constructor of the 's' type, if there is a constructor that accepts
   it/them (determined by overload resolution).

Thanks, guys!
-Howard