[Tex/LaTex] Boolean switch for conditional statement after its definition

conditionals

I'd like to use conditional statements in my preamble file and then define the value of the Boolean variable afterwards. The reason is that I import the preamble.tex first thing in my main file where I then want to specify switches as appropriate. I've tried this with the plain tex \newif and with the etoolbox packages, yet both seem to require the value of the Boolean variable before the definition of the conditional. This appears to mean that I would have to change my preamble.tex, defeating the purpose of making it a separate file.

Is there an easy way of achieving this, short of splitting the preamble file in two parts?

Thanks,
Michael

Best Answer

The approach you're following is not correct. The best thing is to change preamble.tex into preamble.sty and use options.

File `preamble.sty`

\ProvidesPackage{preamble}
\RequirePackage{kvoptions}

% The following defines \ifpreamble@foo (\if<packagename>@<optionname>)
\DeclareBoolOption[true]{foo} % specifying `foo` is the same as `foo=true`

\ExecuteOptions{foo=false} % foo is false by default

\ProcessKeyvalOptions{preamble}

\RequirePackage{fancyhdr}
\pagestyle{fancy}
\fancyhf{}

\ifpreamble@foo
\fancyhead[R]{Michael's report}
\fi

\endinput

File test.tex (`foo` enabled)

\documentclass{article}
\usepackage[foo]{preamble} % or \usepackage[foo=true]{preamble}
\usepackage{kantlipsum}
\begin{document}
\section{A}
\kant
\end{document}

File test.tex (`foo` disabled)

\documentclass{article}
\usepackage{preamble} % or \usepackage[foo=false]{preamble}
\usepackage{kantlipsum}
\begin{document}
\section{A}
\kant
\end{document}

Related Solutions

[Tex/LaTex] etoolbox: environment hooks in boolean switch

You check the boolean switch once in the preamble and make the definition, which is not changed later.

It would work as expected if you would make the test later, such as:

\BeforeBeginEnvironment{ExampleVerbatim}{\ifbool{tightspace}{}{\begin{spacing}{0.5}}}%
\AfterEndEnvironment{ExampleVerbatim}{\ifbool{tightspace}{}{\end{spacing}}}%

However, I would make the spacing value a variable instead. And I guess you meant your tightspace check the other way round as in the original code.

[Tex/LaTex] Using TikZ conditional statement

Welcome! Are you looking for iseven?

\documentclass[tikz]{standalone}

\begin{document}
  \begin{tikzpicture}
   \def\bigdot{5pt}
   \def\littledot{1pt}
    \foreach \x in {0,...,5}{
      \draw (\x,0) circle [radius={iseven(\x)?\bigdot:\littledot}];
    }
  \end{tikzpicture}
\end{document}

In your code you are asking to make the circles for which int(\x/2) is zero big. These are the circles at \x=0 and \x=1, and this is what you get. However, if you want to have "the first, third and fifth dots to be larger" in a list {0,...,5} then you can make the dots larger for which \x is even.

ADDENDUM: As for the different question raised in the comments, you can use mod, as suggested by AlexG, or Mod, which always returns nonnegative values and helps you to avoid confusion (at least I occasionally wasted a lot of time because I was using mod). Both versions are described on p. 1033 of pgfmanual v3.1.5. As above, I prefer the non-deprecated syntax

 circle[radius=<radius>]

over the older, deprecated syntax

 circle(<radius>)

so the proposal for the question in the comments could be

\documentclass[tikz]{standalone}

\begin{document}
  \begin{tikzpicture}
   \path (0,0) node[circle,inner sep=5cm] (c){};
   \def\bigdot{5pt}
   \def\littledot{1pt}
    \foreach \x in {0,...,359}{
      \draw (c.\x) circle [radius={Mod(\x,5)==0?\bigdot:\littledot}];
    }
  \end{tikzpicture}
\end{document}

Note also that if you use \documentclass[tikz]{standalone} then tikz gets automatically loaded, so \usepackage{tikz} is unnecessary. Also, I generally do not really like \defs but if you want to use them, use them locally, i.e. inside the tikzpicture, as above. Yet personally I'd use something like

\documentclass[tikz]{standalone}

\begin{document}
  \begin{tikzpicture}[declare function={rsmall=1;rbig=5;}]
   \path (0,0) node[circle,inner sep=5cm] (c){};
    \foreach \x in {0,...,359}{
      \draw (c.\x) circle [radius={(Mod(\x,5)==0?rbig:rsmall)*1pt}];
    }
  \end{tikzpicture}
\end{document}

Best Answer

File preamble.sty

File test.tex (foo enabled)

File test.tex (foo disabled)