Re: writing MyClass::Operator unsigned long()

From:
"CodeCracker" <sanjaym365@gmail.com>
Newsgroups:
comp.lang.c++
Date:
9 Jun 2006 03:54:52 -0700
Message-ID:
<1149850491.910021.255480@f6g2000cwb.googlegroups.com>
Thanks guys for all the responses.
One more point I would like to know is that how come the function name
have a space in between. How compiler handles this. OR is this a
special case and I can't have a function with a space in between.
Thanks in advance.
CC

Howard wrote:

"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

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