[Tex/LaTex] newcommand with variable number of arguments

conditionalsmacrosparameters

I have seen several related questions, but none seems to address specifically this.

I would like to define a command that accepts a variable number of arguments. Something like

\newcommand{\func}(1){\if{#1}{f(#1)}{f}}

So that if there is a parameter the output will be f(#1) and if there is no parameter the output will be f, and never f().

Is this possible?

Best Answer

The xparse package allows for some really cool syntax stuff.

\documentclass{article}
\usepackage{xparse}

\NewDocumentCommand\funcF{d()}{%
  \IfValueTF{#1}{f(#1)}{f}}

% This is a version that follows more popular LaTeX syntax conventions.
\NewDocumentCommand\NormalFuncF{o}{%
  \IfValueTF{#1}{f(#1)}{f}}

\begin{document}
\[ \funcF(2) = 4       \]
\[ \funcF              \]
\[ \NormalFuncF[2] = 4 \]
\[ \NormalFuncF        \]
\end{document}

output

Related Solutions

[Tex/LaTex] Command with variable number of arguments to format menu sequences

Use menukeys … I guess it’s what your looking for ;-)

\documentclass{article}

\usepackage{menukeys}

% create a new simple style to add arrwos between the items
\newmenustylesimple*{arrows}{\CurrentMenuElement}[ $\to$ ]{blacknwhite}

\begin{document}
\menu{foo > bar > baz}

\bigskip

% apply the new style to the old macro. The optional arguments
% defines the separator, the default is a comma
\renewmenumacro{\menu}[>]{arrows}
\menu{foo > bar > baz}
\end{document}

menukeys

The first line shows the default style of \menu, but it’s possible to change it in many ways. The example shows a very simple style. More predefined styles can be found in the manual …

[Tex/LaTex] Commands that may take a variable number of arguments

As commented, here a solution that uses \@ifnextchar. I also implemented checks against too many or too few arguments (or why are they provided by the user?).

The \@ifnextchar(or its “very internal” big brother \kernel@ifnextchar) skips spaces which results in removed spaces in the third and fourth example.

Code

\documentclass{report}
\usepackage{etoolbox}
\makeatletter
\newcommand*{\decl}[3]{%
  % #1: decl id
  % #2: decl symbol
  % #3: # params
  \csdef{decl@symbol@#1}{#2}%
  \expandafter\newcount\csname c@decl@params@#1\endcsname
  \csuse{c@decl@params@#1}=#3\relax
}
\newcount\decl@params@check
\newcommand*{\usedecl}[1]{%
  \def\decl@name{#1}%
  \edef\decl@params{\the\csuse{c@decl@params@#1}}%
  \def\decl@symbol{\csuse{decl@symbol@#1}}%
  \decl@params@check=\z@
  \let\decl@list\@gobble % the \@gobble removes the first , (expandable)
  \def\decl@next{\kernel@ifnextchar\bgroup\use@decl\use@decl@finish}%
  \decl@next
}
\newcommand*{\use@decl}[1]{%
  \advance\decl@params@check\@ne
  \expandafter\ifnum\the\decl@params@check>\decl@params\relax % too many!
    \PackageWarning{decl}{You have used more params than the \decl@name\space function expected!
                                                    I ignore this (and any following) param, ok?}% but insert the extra argument anyway?!
     \def\decl@next{\use@decl@finish{#1}}% the extra pair of braces {} keeps '#1' local as it is in the input stream
  \else
    \expandafter\def\expandafter\decl@list\expandafter{\decl@list\decl@list@sep#1}%
  \fi
  \decl@next
}
\newif\ifuse@decl@message
\newcommand*{\use@decl@finish}{%
  \ifnum\decl@params@check<\decl@params\relax % too few!
    \expandafter\@firstoftwo
  \else
    \expandafter\@secondoftwo
  \fi
  {%
    \ifuse@decl@message\else
      \PackageWarning{decl}{You have used fewer params than the \decl@name\space function expected! I'm filling up with '??'!}%
      \use@decl@messagetrue
    \fi
    \use@decl{??}}
  {%
    \decl@symbol\decl@list@start\decl@list\decl@list@end
    \use@decl@messagefalse
  }%
}
\newcommand*{\setdeclstart}[1]{\def\decl@list@start{#1}}
\newcommand*{\setdeclend}[1]{\def\decl@list@end{#1}}
\newcommand*{\setdeclsep}[1]{\def\decl@list@sep{#1}}
\makeatother

\setdeclstart{(}
\setdeclend{)}
\setdeclsep{, }

% declare some interface routines
\decl{foo}{FOO}{3}
\decl{bar}{BAR}{4}

\begin{document}
given $P$, $Q$ and $R$ such \dots\ that $\usedecl{foo}{P}{Q}{R}$ results in \dots\par
given P, Q and R such \dots\ that \usedecl{foo}{P}{Q}{R}\ results in \dots\par
\usedecl{foo}{p1}{p2}{p3}\par
\usedecl{bar}{p1}{p2}{p3}{p4}\par
\usedecl{bar}{p1} foo\par
\usedecl{foo}{p1}{p2}{p3} {p4}\par
\end{document}

Output

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