# \newcommand with options

My question is about declaring a "newcommand" with an option.

I have defined

\newcommand{\bforall}[2]{\forall #1\, (#2)}


so that $\bforall{x}{x \in A}$ writes $\forall x (x \in A)$.

I would like some control over the brackets. Namely, I would like

\bforall[s]{x}{x \in A}


to use square braces, and

\bforall[c]{x}{x \in A}


to use curly braces.

How can this be accomplished?

I've search the forum and found 1/ Different command definitions with and without optional argument and 2/ Optional argument for newcommand? and neither helped me.

Here is a way, without packages.

\documentclass{article}
\newcommand\bforall[3][r]{%
\ifx r#1\forall #2\, (#3)\else
\ifx s#1\forall #2\, [#3]\else
\ifx c#1\forall #2\, \{#3\}\else
\mathrm{Illegal~option}%
\fi\fi\fi
}
\begin{document}
$\bforall{x}{x \in A}$

$\bforall[r]{x}{x \in A}$

$\bforall[s]{x}{x \in A}$

$\bforall[c]{x}{x \in A}$

\end{document}


As touhami rightly points out, the penalty for using the simplicity of the \ifx construct is that an accidental syntax error such as using a two character optional argument like [rs] will not produce an error message, but rather just wrongly set the term.

• see @egreg comment to my answer tex.stackexchange.com/questions/254158/… Aug 4, 2015 at 3:12
• @touhami I see the point, but I'm not sure if it applies, since my optional argument has a default non-empty value. To run into the problem, one would have to pass a blank optional argument, I think. But it certainly doesn't hurt to add a \relax after #1 to address that case. In fact, I just tried the case, and it throws the expected error already. Aug 4, 2015 at 9:50
• Sorry, i mean @egreg's last comment, in this case $\bforall[rs]{x}{x \in A}$ Aug 4, 2015 at 10:04
• @touhami In your case, the optional argument was a mathematical term that could be one or more characters. In my case, the optional argument is a single letter indicating round, square, or curly braces. A 2-character optional argument in my case is indicative of a syntax error. Aug 4, 2015 at 10:12
• here is the prolem $\bforall[rs]{x}{x \in A}$ will not idicate error. Aug 4, 2015 at 10:46

Here is one way to do that using the xstring package's \IfStrEqCase:

The default option is round braces, curly braces and square braces. You can also add an optional second parameter to control the brace size:

## Code:

\documentclass{article}
\usepackage{xstring}

\newcommand{\bforall}[3][r]{%
\forall #2\,
\IfStrEqCase{#1}{%
{r}{(#3)}%
{c}{\{#3\}}%
{s}{{[#3]}}%
}[%
%% Issue error message here about unsuported bracket type
]%
}

\begin{document}
$\bforall{x}{x \in A}$

$\bforall[r]{x}{x \in A}$

$\bforall[c]{x}{x \in A}$

$\bforall[s]{x}{x \in A}$

\end{document}


## Code Resizing Brackets:

\documentclass{article}
\usepackage{xstring}
\usepackage{xparse}
\usepackage{amsmath}

\NewDocumentCommand{\bforall}{%
O{r}% #1 = r, c or s (bracket shape)
O{}%  #2 = optional size specifier
m%    #3
m%    #4
}{%
\forall #3\,
\IfStrEqCase{#1}{%
{r}{#2(#4#2)}%
{c}{#2\{#4#2\}}%
{s}{{#2[#4#2]}}%
}[%
%% Issue error message here about unsupported bracket type
\PackageError{tree}{Undefined option to tree: #1}{}
]%
}

\begin{document}
\begin{minipage}{0.25\linewidth}
$\bforall{x}{x \in A}$

$\bforall[r]{x}{x \in A}$

$\bforall[c]{x}{x \in A}$

$\bforall[s]{x}{x \in A}$
\end{minipage}
\hspace*{0.5cm}
\begin{minipage}{0.25\linewidth}
$\bforall[r][\Big]{x}{x \in A, x > \dfrac{1}{2}}$

$\bforall[c][\Big]{x}{x \in A, x > \dfrac{1}{2}}$

$\bforall[s][\bigg]{x}{x \in A, x > \dfrac{1}{2}}$
\end{minipage}
\end{document}


For a case switch macro the best is expl3; I used the letters already common with amsmath, p for parentheses, b for brackets, B for braces.

\documentclass{article}
\usepackage{amsmath}
\usepackage{xparse}

\ExplSyntaxOn

\NewDocumentCommand{\bforall}{O{p}mm}
{
\forall #2\,
\str_case:nnF { #1 }
{
{p}{(#3)}
{b}{[#3]}
{B}{\{#3\}}
}
{\@latex@error{Illegal~option~#1}{I~used~p}}
}

\ExplSyntaxOff

\begin{document}

$\bforall{x}{x \in A}$

$\bforall[p]{x}{x \in A}$

$\bforall[b]{x}{x \in A}$

$\bforall[B]{x}{x \in A}$

\end{document}


A plain TeX one with a different syntax

\def\bforall#1{\forall#1\,\futurelet\tmptoken\dobforall}
\def\dobforall{\ifx\tmptoken\bgroup\expandafter\bforallbraces\fi
\ifx\tmptoken[\expandafter\bforallbrackets\fi
\ifx\tmptoken(\expandafter\bforallparenthesis\fi
\relax}
\def\bforallbraces#1\relax#2{\{#2\}}
\def\bforallbrackets#1\relax[#2]{[#2]}
\def\bforallparenthesis#1\relax(#2){(#2)}

$\bforall{x}(x \in A)$\par
$\bforall{x}{x \in A}$\par
$\bforall{x}[x \in A]$\par

\bye


You can use those definitions also in LaTeX without the need of any package. The syntax seems easier for me \bforall{x}(x\in A), \bforall{x}{x\in A} and \bforall{x}[x\in A]. In any case, for your needs, may be you don't need that macro at all, just \def\bforall#1{\forall#1\,} and then it's followed by whatever you input?

With a little bit of refinement to let you use \bforall x {x \in A} syntax. (In LaTeX it would be easier because of \@ifnextchar.)

\def\bforall#1{\forall#1\,\futurelet\tmptoken\dobforall}
\def\dobforall{\ifx\tmptoken\spacetoken\expandafter\bforallspace\fi
\ifx\tmptoken\bgroup\expandafter\bforallbraces\fi
\ifx\tmptoken[\expandafter\bforallbrackets\fi
\ifx\tmptoken(\expandafter\bforallparenthesis\fi
\relax}
\def\bforallbraces#1\relax#2{\{#2\}}
\def\bforallbrackets#1\relax[#2]{[#2]}
\def\bforallparenthesis#1\relax(#2){(#2)}
\def\bforallspace#1\relax{\dobforallspace}
\expandafter\def\expandafter\dobforallspace\space{\futurelet\tmptoken\dobforall}
\lowercase{\let\spacetoken= } %

$\bforall x (x \in A)$\par
$\bforall x {x \in A}$\par
$\bforall x [x \in A]$\par

\bye