Is the following code valid?

void foo()
{
    int i;
    typeid( i );
}

I don't have typeinfo included, so I don't know
what std::type_info is. However, I'm just throwing
the result of typeid away.



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5.2.8 p6
"If the header <typeinfo> (18.5.1) is not included prior to a use of typeid,
the program is ill-formed."

Is that what your wondering? ..

--
Algorithms
http://www.cs.ircnetworks.com

"Anon" <spamme@lots.moderated.news.pipex.net> wrote in message
news:3a8c8cb8_1@news.nwlink.com...
| Is the following code valid?
|
| void foo()
| {
|     int i;
|     typeid( i );
| }
|
| I don't have typeinfo included, so I don't know
| what std::type_info is. However, I'm just throwing
| the result of typeid away.
|
|
|
| ---
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|

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Anon wrote:
>
> Is the following code valid?
>
> void foo()
> {
>     int i;
>     typeid( i );
> }
>
> I don't have typeinfo included, so I don't know
> what std::type_info is. However, I'm just throwing
> the result of typeid away.

All you need to know is that typeid() returns a value. C++ always allows
you to ignore a returned value. As long as the typeid expression itself
is valid, and it seems to be, this should work. Did you remember to
#include <typeinfo>?

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>===== Original Message From "Anon" <spamme@lots.moderated.news.pipex.net>
=====
>Is the following code valid?
>
>void foo()
>{
>    int i;
>    typeid( i );
>}
>
>I don't have typeinfo included, so I don't know
>what std::type_info is. However, I'm just throwing
>the result of typeid away.

It doesn't matter what you do or don't do with the result.  The
Standard is clear in 5.2.8p6:

        If the header <typeinfo> (18.5.1) is not included prior to a
        use of typeid, the program is ill-formed.

Your example is a use of typeid, even though the result is ignored.

-- William M. Miller

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