🚀 go-pugleaf

In article <b2dt3k$2g0n$1@news.dataguard.no>, Conrad Weyns
<weyns@online.no> wrote:

| "Miro Jurisic" <macdev@meeroh.org> skrev i melding
| news:macdev-B2A383.10035512022003@senator-bedfellow.mit.edu...
| > In article <b2dg2g$1117$1@news.dataguard.no>, "Conrad Weyns"
| <weyns@online.no>
| > wrote:
| >
| > > compiler: CW Pro 8.3 Mac & Win.
| > > issue: how can I get hold of the real type name of T that is part of a
| > > function signature in the following stripped to the bone but working
| code:
| > >
| > > Output of this running program:
| > >
| > > 1A. tA.report() gives:
| > >    function name: _report<long>(long &)
| > >    using __typeof__: long
| > >    sWoes address: 0x00420460 sWoes = 999
| > >    type of _t: long
| > >    value of _t 123
| > >
| > > 2A. tA.const_report() gives:
| > >    function name: _report<const long>(const long &)
| > >    using __typeof__: long
| > >    sWoes address: 0x00420454 sWoes = 999
| > >    type of _t: long
| > >    value of _t 123
| >
| > My understanding is that typeof (T) == typeof (const T) == typeof
| (volatile T)
| > == etc. Perhaps if you explain exactly what you are trying to do with this
| > information we can come up with an answer. (Which will involve type
| traits, I am
| > sure.)
| 
| Thanks meeroh, looks like I need to go back to school -again :-(
| Ehm, I am lying now, I never went to school!
| 
| But this was the kick in the behind that I needed to get to an alternative
| solution. I can simply differentiate between the two cases by supplying
| *two* template functions:
| 
| template <typename T> void _report(T& _t);
| template <typename T> void _report(const T& _t);
| 
| Works for me :-)

Actually the semantics of typeof haven't been worked out by the C++
committee yet.  But what meeroh says is certainly true for our current
implementation of it.

Glad you found a solution!

If it helps any further, there is a neat little type traits (hinted at
by meeroh) that lives in <msl_utility>.  You could have stuck it in
your original _report like so:

    if (Metrowerks::is_const<T>::value)
      cout << "const\n";

And it would have printed "const" for the const variants.

-- 
Howard Hinnant
Metrowerks