Re: composition of declarators

From:

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

Newsgroups:

comp.lang.c++

Date:

Tue, 24 Apr 2007 09:20:07 -0400

Message-ID:

<f0l068$ao$1@news.datemas.de>

Sylvester Hesp wrote:

"Victor Bazarov" <v.Abazarov@comAcast.net> wrote in message
news:FcOdnS1xnNYD7rDbnZ2dnUVZ_g-dnZ2d@comcast.com...

ssailor wrote:

I saw an exaple in the c++ standard(clause 6.8),

-------------------------------------------------------
struct T1 {
     T1 operator ()( int x ) { return T1(x ); }
     int operator =( int x ) { return x; }
     T1(int ) { }
};
struct T2 { T2(int ){ } };
int a , (*(* b)( T2 ))( int ), c , d;
void ff ()
{
      T1(a) = 3 ,
      (*(* b)( T2(c )))( int(d )); // AA
}
--------------------------------------------

I am confused that the declaration at the line AA is valid.

It is.

Then, what's the meaning of this declaration? Thank in advance.

The second part of the declaration declares 'b' to be a pointer
to a function that takes one argument of type 'int' and returns
a pointer to a function that takes one argument of type T2 and
returns T1.

Actually, it is the other way around: it is a function pointer taking
a T2 and returning a pointer to a function taking an int and
returning a T1.

Right. My mistake.

The funny thing is, if you replace the comma on the previous line
with a semi-colon, it actually becomes a double function call: first
on the function pointed to by the earlier defined 'b', passing a T2
constructed out of c, and another function call on the function
returned by b(), passing an int constructed out of d)

The funnier thing is, if you drop 'T1' from the line above, the
entire statement becomes executable, assigning 3 to 'a' and then
calling 'b' and the other function as you described.

V
--
Please remove capital 'A's when replying by e-mail
I do not respond to top-posted replies, please don't ask