Why does std.variant not have a tag?

[Alex Rønne Petersen]

On 05-11-2012 00:31, evansl wrote:
>    http://dlang.org/phobos/std_variant.html
>
> says:
>
>   This module implements a discriminated union type (a.k.a. tagged union,
> algebraic type).
>
> Yet, the wiki page:
>
>    http://en.wikipedia.org/wiki/Tagged_union
>
> says:
>
>    a tag field explicitly indicates which one is in use.
>
> and I don't see any indication of a tag field in the std_variant.html
> page.  Another wiki reference:
>
>    http://en.wikipedia.org/wiki/Disjoint_union
>
> is more explicit because it pairs the tag with the value:
>
>    (x,i)
>
> where x is the value and i is the tag.
>
> One reason for an explicit tag is that the bounded types may contain
> the same type twice.  This has lead to problems in boost::variant as
> evidenced by the post:
>
>    http://article.gmane.org/gmane.comp.parsers.spirit.general/17118
>
> In addition, both variant and tuple have a common part, a metafunction
> mapping from a tag to a type; hence, this same common part could be used
> to implement both tuple and a tagged variant.
>
> A variant which actually contained a tag field I think would be more
> general in that it would allow duplicate types among the bounded types
> just as a tuple's bounded types can contain duplicate types.
>
> -regards,
> Larry
>

Yes, this is a big problem with the current std.variant implementation 
(among other problems such as no recursive variants....). The best 
std.variant can offer right now is the 'type' property to identify 
what's stored in it.

std.variant is, unfortunately, not very useful if you want the semantics 
of variants in ML-style languages.

-- 
Alex Rønne Petersen
alex at lycus.org
http://lycus.org