Re: Enum oddity

From:
Francesco <entuland@gmail.com>
Newsgroups:
comp.lang.c++
Date:
Tue, 25 Aug 2009 07:44:22 -0700 (PDT)
Message-ID:
<62e84e27-222e-499a-8459-75c064b75f7a@v20g2000yqm.googlegroups.com>
On 25 Ago, 16:28, Victor Bazarov <v.Abaza...@comAcast.net> wrote:

Francesco wrote:

On 25 Ago, 11:44, James Kanze <james.ka...@gmail.com> wrote:

[..]
The case I run into the most is when I want to initialize the
array directly, instead of using copy:


                    ^^^^^^^

    std::vector< int > my_vector(
        std::istream_iterator< int >( std::cin ),
        std::istream_iterator< int >() ) ;

Stupid compiler doesn't realize that if I'm naming it
"my_vector", then I don't want a function. (And I use this
idiom a lot, since if I'm not modifying the vector later, it
allows me to declare it const.)


Sorry for replying to a post which wasn't addressed to me, James, but
the code you posted above compiles and runs fine (it creates a vector
instead of declaring a function) - at least on my environment.


What environment is that? And how do you know it creates a vector?

 > Is it

really expected to rise the parsing problem?


Parsing problem? In what way? Try compiling this:

#include <vector>
#include <iostream>
#include <iterator>
#include <istream>

int main()
{
     std::vector< int > my_vector(
         std::istream_iterator< int >( std::cin ),
         std::istream_iterator< int >() ) ;

     my_vector.size();

}

It's not the declaration of 'my_vector' itself that doesn't compile.
The declaration is not of an object, though. The compiler will choke
when I try using the dot operator to call the 'size' member function of
'std::vector'. And the reason is simple: 'my_vector' is not an object
but a function.


Victor, I'm taking on the habit to test the code before making my
assertions. I believe you didn't test your code, because no compiler
should accept declaring a parameter function called "std::cin". In
fact, it doesn't interpret that as a function declaration but as a
vector instantiation. Try that yourself.

Francesco

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