0

See the code that follows.

The user gives some pairs of numbers. Each time they happen also to be displayed.

Finally they are added up in a specific way.

\documentclass{article}

%\def\Result...

\newcount{\A}\A=0

\newcount{\B}\B=0

\newcount{\C}\C=0

\def\XYZ{#1,#2}{\textbf{#1}---\textbf{#2}\A=#1\B=#2\multiply\A by 100\advance\C by \A\advance\C by \B}

\begin{document}
\XYZ{4,30}

\XYZ{7,30}

\XYZ{9,40}

%\Result

\end{document}

This kind of code prevents the document from ever compiling.

  1. What is wrong with it? How can it be fixed?
  2. Can it be rewritten using \newcounter, \setcounter, \countainedwithin*, and other LaTeX methods? I am interested in knowing both the correct TeX and LaTeX methods to achieve the result.
  3. How does one write a macro \Result that takes \C and divides by 100 and if the result is three digits EFG, writes \textbf{E},\textbf{FG}, and if the result is four digits EFGH, writes \textbf{EF},\textbf{GH} instead?
  4. In case it has fewer than three digits, to add 0 for G or 00 for FG and use the three digit case.

I am looking for something that does not involve the calc package by the way. A result rounded to the nearest FG or GH in \Result is also fine.

For the \Result 

\def\Result{\divide\C by 100\textbf{\the\C}}

is working now - but now sure how to handle rounding up to nearest H in the EF.GH.

If it did work, not sure how to handle the fractional part and cut off to at most two digits on each side of a dividing marker.

Improve this question
10
  • It does not combine independent of what \Result is defined as being. So the first problem is in the counters.
    – Gottfried William
    Apr 6, 2020 at 21:45
  • Also the issue may be in using {#1,#2} in the definition. However it is not ideal to have to enter pairs by \XYZ{X,Y.} in the use document, if doing something like \def\XYZ#1,#2.{...
    – Gottfried William
    Apr 6, 2020 at 21:50
  • 1
    \def\XYZ{#1,#2} is a syntax x error.
    – David Carlisle
    Apr 6, 2020 at 21:53
  • 1
    for part 3 of your question \C / 100 in this example is 21 so neither three nor 4 digits, what output do you want in this case?
    – David Carlisle
    Apr 6, 2020 at 22:07
  • 1
    Usual syntax for \newcount is to give the argument without braces (it's a plain TeX macro). Not that it doesn't work with the braces, but I find it a bit strange (if you were to play this game with primitives such as \let or \ifx, you'd get bitten).
    – frougon
    Apr 7, 2020 at 0:13

1 Answer 1

Reset to default
2

It's not at all clear what calculation or output you want so I fixed the syntax errors, left the calculation, then made it typeset the values of A B and C

enter image description here

\documentclass{article}

%\def\Result...

\newcount{\A}\A=0

\newcount{\B}\B=0

\newcount{\C}\C=0

\def\XYZ#1{\XYZa#1\relax}
\def\XYZa#1,#2\relax{\textbf{#1}:\textbf{#2}%
\A=#1\relax
\B=#2\relax
\multiply\A by 100\relax
\advance\C by \A
\advance\C by \B
[\the\A,\the\B,\the\C]%
}

\begin{document}
\XYZ{4,30}

\XYZ{7,30}

\XYZ{9,40}

%\Result

\end{document}
Improve this answer
2
  • I see, \relax is the "magic" in these cases.
    – Gottfried William
    Apr 6, 2020 at 22:01
  • 1
    @GottfriedWilliam not really: mostly they are not needed but you had not terminated any of the assignments \A=#1 would set \A to be #1 followed by whatever digits the following tokens expand to, you were just lucky that the following tokens don't expand to digits here. Using \relax makes the end of the assignment explicit. The main change is fixing the \def syntax and making it output some text.
    – David Carlisle
    Apr 6, 2020 at 22:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.