On Tuesday, 30 July 2013 at 20:34:32 UTC, Ali Çehreli wrote:
On 07/30/2013 12:09 PM, JS wrote:
> I already stated why this is not a proper example, I'm not
using Pragma
> in run time code(for lack of a better term).
>
> module main;
>
> import std.stdio;
>
>
> template Pragma(alias amsg)
> {
> void Pragma(string file = __FILE__)
> {
> pragma(msg, amsg);
> }
> }
>
> template t()
> {
> Pragma!("help, this does not work!!!!!!!!!");
> }
>
> void main()
> {
> enum msg = "hello";
> Pragma!msg;
> t!();
> }
Thank you. Now we have something to work on. The program above
produces the following error:
Error: no identifier for declarator Pragma!"help, this does not
work!!!!!!!!!"
The error is fixed by adding a mixin:
mixin Pragma!("help, this does not work!!!!!!!!!");
Or just assigning it to enum... both are not great solutions...
Despite another error it actually works:
hello
help, this does not work!!!!!!!!! <-- IT WORKED
Error: t!() has no effect
The second error is fixed by another mixin:
mixin t!();
Perhaps the example is too simplistic. Still, let's stay with
it for further issues.
Ali
I think that such behavior should not be an error but possibly a
warning, if at all.
And if you cared to read what I initially posted you would
realize I already explained the same thing. The example maybe
more clear but I stated, maybe in some convoluted and jumped way,
that Pragma doesn't work *in* templates..
"When I try to use void instead of string and do something like
Pragma!(msg)
I get an error that the template has no effect."
and later
"It's goal is to debug templates and essentially
just wrapping pragma to supply the __FILE__ info automatically.
e.g., instead of having to do
pragma(msg, __FILE__~amsg);
I want to do Pragma!(amsg);
"
