Re: \dorecurse and \recurselevel in Natural Tables

[Aditya Mahajan <adityam@umich.edu>]

On Fri, 6 Mar 2009, Wolfgang Schuster wrote:

>
> Am 06.03.2009 um 04:05 schrieb Aditya Mahajan:
>
>> My summary of this thread:
>> http://randomdeterminism.wordpress.com/2009/03/05/tex-programming-the-past-the-present-and-the-future/
>
> Nice article but I have a few corrections for you.
>
>
> Table setup:
>
> To achieve horizontal and vertical centered content in the table cell
> you need 'align={middle,lohi}'.

Hmm... align={middle,middle} also works. I haven't checked the code to see 
why.

>
> Pure table code:
>
> You forgot the \bTR and \eTR before and after each row.

Thanks.

>
> Expandafter solution:
>
> You don't need as many \expandafter as you did in your example,
> one before \bTD in each loop is enough.
>
> \bTABLE
> \bTR
>  \bTD $(+)$ \eTD
>  \dorecurse{6}
>   {\expandafter \bTD \recurselevel \eTD}
>  \eTR
> \dorecurse{6}
> {\bTR
>     \edef\firstrecurselevel{\recurselevel}
>     \expandafter\bTD \recurselevel \eTD
>  \dorecurse{6}
>    {\expandafter
>     \bTD
>      \the\numexpr\firstrecurselevel+\recurselevel\relax
>     \eTD}
>  \eTR}
> \eTABLE

Ah. I was experimenting without the \the\numexpr (replacing both by 
\relax) so that I could see the output. I did not realize that \numexpr 
expands its arguments anyways.
>
> \for loop:
>
> If you really want to show this loop you also the correct
> commands to access the counters. #1 did also work but only
> because the command is wrapped around \dostepwiserecurse.
>
> \bTABLE
>  \bTR
>    \bTD $(+)$ \eTD
>    \for \y=1 \to 6 \step 1 \do
>      {\bTD \y \eTD}
>  \eTR
>  \for \x=1 \to 6 \step 1 \do
>  {\bTR
>     \bTD \x \eTD
>    \for \y=1 \to 6 \step 1 \do
>    {\bTD \the\numexpr\x+\y \eTD}
>  \eTR}
> \eTABLE

This does not work, for the same reason that the \dorecurse solution does 
not work. You need to exapnad \x and \y in the first row and first column.

> Thank you for the article and I'm looking forward for the next.

Thanks for the feedback.

Aditya
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