Re: [BUG] debug.getlocal does not pick up vararg

[Dirk Laurie <dirk.laurie@...>]

2016-07-26 14:03 GMT+02:00 Luiz Henrique de Figueiredo <lhf@tecgraf.puc-rio.br>:
> Here is a similar script that is easier to interpret:
>
> function f(a,b,...) for k=-4,4 do print(k,debug.getlocal(1,k)) end end
> f(10,20,30,40,50)
>
> The output is below.
> Your complaint is that 5.3.2 no longer shows "(*vararg)" and their values?

Yes. It is totally redundant to say "Negative indices refer to vararg
parameters; -1 is the first vararg parameter." when exactly the same
result is returned whether or not the parameter is actually present.