Re: writing MyClass::Operator unsigned long()

From:

"Victor Bazarov" <v.Abazarov@comAcast.net>

Newsgroups:

comp.lang.c++

Date:

Thu, 8 Jun 2006 13:10:26 -0400

Message-ID:

<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'.

The part '= s()' is called an initializer, and the process that involves
the whole thing is called "copy-initialization". Another syntax for that
is

   unsigned long ul((s()));

(notice double parentheses around 's()', they are required because this:

   unsigned long ul(s());

is a function declaration (but that's a different story).

(*) 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).

V
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