Why is "like result" not allowed?

Is there any problem with using it?
Spec says only final feature name or "current".

In message <293598cc-bcfb-433a-b683-d0bc5b161f5f@s21g2000prm.googlegroups.com>
scholz.lothar@gmail.com wrote:

> Is there any problem with using it?
> Spec says only final feature name or "current".

Isn't it pointless?

The declaration would have to be a local to a routine, which will know its
own result type, and to change the routine's result type you'd have to write
another routine anyway.

Simon

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