Re: how can i short name a compile time state ?

From:
James Kanze <james.kanze@gmail.com>
Newsgroups:
comp.lang.c++
Date:
Mon, 25 Feb 2008 00:56:07 -0800 (PST)
Message-ID:
<a1f0870a-7f23-408f-9320-cd807efb12ba@v3g2000hsc.googlegroups.com>
On Feb 25, 8:36 am, toton <abirba...@gmail.com> wrote:

  I want to remember a compile time variable state, just to have some
typing convenience. I am not sure if I can do it using some typedef.
To give a short example,
I have an enum as,

enum dir_type{
  dir_x,dir_y,dir_xd
};

and 2 specializations as,

template<dir_type>
dir_type o_dir();

template<>
dir_type o_dir<dir_x>(){
        return dir_y;}

template<>
dir_type o_dir<dir_y>(){
        return dir_x;
}

now in a function I want to do,

template<dir_type dt>
void funct(){
        std::cout<<dt<<std::endl;
        dir_type od = o_dir<dt>(); /// what would be compile time equiv =

of

this line so than i can call next line?
        dir_type d = o_dir<od>();}

Now for the commented line, I want a compile time typedef , so
that od is a name rather than a runtime state, and I can call
the next line. This is just to remove repetitive writing of
o_dir<dt> , in several places when i can simply write od.


As you've written it, you can't. od is a value, initialized by
the return value of a function, and as such, is not a compile
time constant (at least not yet---there's some discussion of
allowing such trivial functions to be used in constant
expressions).

I'm not too sure what you're trying to achieve. Perhaps
something like:

    template< dir_type >
    struct o_dir ;
    template<>
    struct o_dir< dir_x >
    {
        static dir_type const other = dir_y ;
    } ;
    // etc.

could be used, e.g.:

    template< dir_type dt >
    void
    funct()
    {
        dir_type const od = o_dir::other ;
        dir_type d = o_dir< od >::other ;
        // ...
    }

--
James Kanze (GABI Software) email:james.kanze@gmail.com
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