Is there an idiom, preferably D1, to detect whether or not the code is currently executing as a ctfe?

Ie:

void foo()
{
    (static?) if( ???? )
    {
        // Run-time code here
        // that does stuff the CTFE engine chokes on
    }
    else
    {
        // CTFE code here
        // that might not be ideal, but at least works as CTFE
    }
}

Nick Sabalausky wrote:
> Is there an idiom, preferably D1, to detect whether or not the code is currently executing as a ctfe?
> 
> Ie:
> 
> void foo()
> {
>     (static?) if( ???? )
>     {
>         // Run-time code here
>         // that does stuff the CTFE engine chokes on
>     }
>     else
>     {
>         // CTFE code here
>         // that might not be ideal, but at least works as CTFE
>     }
> }
> 
> 

Not currently, but you can use the following hack, which exploits a bug with CTFE (if the bug gets fixed then the code below won't work):

----
import tango.io.Stdout;

bool isCtfe()
{
        void test( char[] str )
        {
                str[0] = 'b';
        }
        char[] a = "foo".dup;
        test(a);
        if( a == "boo" )
                return false;
        return true;
}

const ctfe = isCtfe();

void main()
{
        Stdout.formatln( "CTFE: {}; Runtime: {};", ctfe, isCtfe() );
}
----

Nick Sabalausky wrote:
> Is there an idiom, preferably D1, to detect whether or not the code is currently executing as a ctfe?
> 
> Ie:
> 
> void foo()
> {
>     (static?) if( ???? )
>     {
>         // Run-time code here
>         // that does stuff the CTFE engine chokes on
>     }
>     else
>     {
>         // CTFE code here
>         // that might not be ideal, but at least works as CTFE
>     }
> }

Don posted one to the NG once, but I'm not sure if it works with D1:

http://www.digitalmars.com/d/archives/digitalmars/D/Going_from_CTFE-land_to_Template-land_101395.html#N101414

-Lars