Re: Template mixin enum stringof

[Lemonfiend via Digitalmars-d-learn]

On Wednesday, 10 December 2014 at 12:57:16 UTC, ketmar via 
Digitalmars-d-learn wrote:
On Wed, 10 Dec 2014 12:35:44 +0000
Lemonfiend via Digitalmars-d-learn 
<digitalmars-d-learn@puremagic.com>
wrote:
On Wednesday, 10 December 2014 at 12:08:34 UTC, ketmar via 
Digitalmars-d-learn wrote:
> On Wed, 10 Dec 2014 11:52:11 +0000
> Lemonfiend via Digitalmars-d-learn 
> <digitalmars-d-learn@puremagic.com>
> wrote:
>
>> Consider the following:
>> 
>> ---
>> enum Foo { BAR, }
>> 
>> mixin template S(string s_)
>> {
>>        enum s = s_;
>> }
>> 
>> void main()
>> {
>> 	mixin S!(Foo.BAR.stringof); // Error: identifier 
>> 'stringof' of 'Foo.BAR.stringof' is not defined
>> 
>> 	enum e = Foo.BAR.stringof;
>>        mixin S!(e); // works fine
>> }
>> ---
>> 
>> Why doesn't the first work? And is there an alternative to 
>> the second version?
>
>   mixin S!((Foo.BAR).stringof);
>
> sorry, don't remember the parsing rule for this.

Wow, I didn't even consider that.. Thanks!
also, you can use this:

  import std.conv : to;
  ...
  mixin S!(to!string(Foo.BAR));

people tend to forget that many `to!` variants are usable in 
CTFE.

Seems like I'd want to move the converting-to-string 
functionality to within the template mixin then, something like:

---
mixin template S(T, T t) if (is(T == enum))
{
        //import std.traits;
	//enum s = fullyQualifiedName!(t); // unfortunately this results 
in "t"

        import std.conv: to;
        enum s = to!string(t); // this works
}

mixin S!(Foo, Foo.BAR);
---

But passing an enum as parameter seems to be somewhat annoying. 
If I leave off the first Foo, then it complains about no-matching 
template for int parameter.
!(Foo, Foo.BAR) seems kinda redundant..