Re: stacking

Aditya Mahajan <[email protected]>
Newsgroups gmane.comp.tex.context
Message-ID <[email protected]>
On Mon, 6 Apr 2026, Emanuel Han wrote:

> Hi list,
> 
> I want to show snippets of text on a slide and then add some other snippets
> of the same text on the next slide. Then, restart with some new text, on a
> „cleared“ slide.
> 
> Example:
> 
> Slide 1: English _______ difficult.
> Slide 2: English   is    difficult.
> Slide 3: ______ English classes ______ Mick ______ fun.
> Slide 4:   The  English classes  with  Mick  are   fun.
> 
> Let’s call „question1“ what is to be printed on slide 1.
> Let’s call „answer1“ what is to be added to that on slide 2.
> Let’s call „question2“ what is to be printed on slide 3.
> Let’s call „answer2“ what is to be added to that on slide 4.
> And so on.
> 
> My first attempt is with stacking.
> 
> It works so far, the only problem I couldn’t solve is how to „clear“ the
> space used up by previous questions (and their answers). Is there a specific
> way to do this with stacking? I was not successful in finding documentation
> about stacking.

You should really think of this as two separate "stacks": one for Q1 and one for Q2. 


\setuppapersize[S6][S6]

\startbuffer[Q1]
\startstackingsteps[1,{1:2}]
   \stacking[1]{First Question}

   \stacking[2]{First Answer}
   \page
 \stopstackingsteps
\stopbuffer

\startbuffer[Q2]
\startstackingsteps[1,{1:2}]
   \stacking[1]{Second Question}

   \stacking[2]{Second Answer}
   \page
 \stopstackingsteps
\stopbuffer

\starttext

\dorecurse{2}{\getbuffer[Q\recurselevel]}

\stoptext

 
> And then I tried to include a counter, so that I can re-arrange the order of
> the questions with just cutting and pasting them somewhere else in the source
> code, without the need to manually edit the number of the question (even
> better than that would be a randomizer for the order of the questions). Below
> is my attempt with the counter, which unfortunately doesn’t work. Maybe it’s
> a matter of when the counter variable is stored / updated.

This should be easy with the above structure.

Aditya

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