General

hkyz · Post by **hkyz** » Mon Feb 11, 2013 6:14 am

I want to overload a definition with three variations using arguments in parenthesis. Something like the non-working code below:

Code: Select all

\documentclass{article}

\def\f{F}
\def\f(#1){F(#1)}
\def\f(#1,#2){F_{#1}^{#2}}

\begin{document}

$\f$
$\f(1)$
$\f(1,2)$

\end{document}

Any ideas on how to do such a thing without too much effort?

Thanks.

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sommerfee · Post by **sommerfee** » Mon Feb 11, 2013 10:11 am

Code: Select all

\documentclass{article}
\begin{document}

\makeatletter
\def\f{\@ifnextchar(\@f F}
\def\@f(#1){\@@f(#1,,)}
\def\@@f(#1,#2,#3){%
  \ifx\relax#2\relax
    F(#1)%
  \else
    F_{#1}^{#2}%
  \fi}
\makeatother

$\f$
$\f(1)$
$\f(1,2)$

\end{document}

For explanation see http://www.ctan.org/pkg/texbytopic

hkyz · Post by **hkyz** » Thu Feb 14, 2013 12:09 am

Thanks for the answer.

LaTeX.org

General ⇒ New Command with Arguments in Parentheses

New Command with Arguments in Parentheses

New Command with Arguments in Parentheses

Re: New Command with Arguments in Parentheses