$-variables capture

[veretin]

Hi!

Suppose, we want to expand each A(q?) as B(q)+C(q) if there is tag(q).
Here is the code with $ variables.

c A,B,C,tag,tag2;
v q1,q2,q3,k;

g ff =
       + A(q1)*A(q2)*A(q3) * tag(q2)*tag(q3);

#do i=1,2
*  repeat;                                                                                                                              
    if ( match( tag(k?$k) ) );
      id tag($k) = tag2($k);
      id,many, A($k) = B($k) + C($k);
    endif;
*  endrepeat;                                                                                                                           
*.sort                                                                                                                                  
#enddo

id tag2(?a) = tag(?a);
moduleoption local $k;
.sort

b tag,tag2,A;
print+s ff;
.end

This gives

      + A(q1)*tag(q2)*tag(q3) * (
          + B(q2)*B(q3)
          + B(q2)*C(q3)
          + B(q3)*C(q3)
          + C(q3)^2
          );

which I find wrong.
But if I uncomment .sort then it works:

 + A(q1)*tag(q2)*tag(q3) * (
          + B(q2)*B(q3)
          + B(q2)*C(q3)
          + B(q3)*C(q2)
          + C(q2)*C(q3)
          );

With repeat, instead of #do-cycle, the same story.

In this particular case we can work without $-variable

id tag(q?)*A(q?) = tag(q)*(B(q) + C(q));

But in real code, I need some further manipulations with q.

[vermaseren]

What happens is that after it generates the first B term, the new term runs into the match and gets a new value for $k. Only when it is done with the second tag and A->B+C it comes back at the first C, but then with the newer value of $k. This is the way it is supposed to work. That it is not what you want means that you have to force it to finish the first A->B+C with the .sort This is not an error. It is a consequence of the model. If you do not like the .sort, another way it to place things inside function arguments in such a way that each argument gets sorted separately. It all depends on how expensive the .sort is in your program.

…

On 2 Jun 2026, at 18:25, veretin ***@***.***> wrote: veretin created an issue (form-dev/form#835) <#835> Hi! Suppose, we want to expand each A(q?) as B(q)+C(q) if there is tag(q). Here is the code with $ variables. c A,B,C,tag,tag2; v q1,q2,q3,k; g ff = + A(q1)*A(q2)*A(q3) * tag(q2)*tag(q3); #do i=1,2 * repeat; if ( match( tag(k?$k) ) ); id tag($k) = tag2($k); id,many, A($k) = B($k) + C($k); endif; * endrepeat; *.sort #enddo id tag2(?a) = tag(?a); moduleoption local $k; .sort b tag,tag2,A; print+s ff; .end This gives + A(q1)*tag(q2)*tag(q3) * ( + B(q2)*B(q3) + B(q2)*C(q3) + B(q3)*C(q3) + C(q3)^2 ); which I find wrong. But if I uncomment .sort then it works: + A(q1)*tag(q2)*tag(q3) * ( + B(q2)*B(q3) + B(q2)*C(q3) + B(q3)*C(q2) + C(q2)*C(q3) ); With repeat, instead of #do-cycle, the same story. In this particular case we can work without $-variable id tag(q?)*A(q?) = tag(q)*(B(q) + C(q)); But in real code, I need some further manipulations with q. — Reply to this email directly, view it on GitHub <#835?email_source=notifications&email_token=ABJPCESK3RPI6PSEDHL6EPL4535X7A5CNFSL4Z3JMQ5C6L3HNF2C22DVMIXUS43TOVSS6NBVG4ZTCNZRGY2TTJTSMVQXG33OVJZXKYTTMNZGSYTFMSSWK5TFNZ2KYZTPN52GK4S7MNWGSY3L>, or unsubscribe <https://github.com/notifications/unsubscribe-auth/ABJPCEWJ3D4NF76ENJPERRD4535X7AVCNFSM6AAAAACZXNSS6OVHI2DSMVQWIX3LMV43ASLTON2WKOZUGU3TGMJXGE3DKOI>. Triage notifications, keep track of coding agent tasks and review pull requests on the go with GitHub Mobile for iOS <https://github.com/notifications/mobile/ios/ABJPCEWNWFTCW34HB2H3DL34535X7A5CNFSL4Z3JMQ5C6L3HNF2C22DVMIXUS43TOVSS6NBVG4ZTCNZRGY2TTJTSMVQXG33OVJZXKYTTMNZGSYTFMSSWK5TFNZ2KUZTPN52GK4S7NFXXG> and Android <https://github.com/notifications/mobile/android/ABJPCEVEZMVUDAH7FNIRPNT4535X7A5CNFSL4Z3JMQ5C6L3HNF2C22DVMIXUS43TOVSS6NBVG4ZTCNZRGY2TTJTSMVQXG33OVJZXKYTTMNZGSYTFMSSWK5TFNZ2K4ZTPN52GK4S7MFXGI4TPNFSA>. Download it today! You are receiving this because you are subscribed to this thread.