\newcommand for a function with one optional argument inside other optional argument

Question

SX. In this case, I want to create a \newcommand named \fun for function.

My problem arises when I use a \ fun command again with the name f (the letter f is widely used in my document) and avoid the first definition because otherwise, I will get the first case.

Minimal example

\documentclass{memoir}
\usepackage{amsmath,amssymb}
\usepackage{xstring}
\newcommand{\rn}[1][n]{%
    \IfEqCase{#1}{%
        {1}{\mathbb{R}}%
        {2}{\mathbb{R}^{2}}%
        {3}{\mathbb{R}^{3}}%
        {n}{\mathbb{R}^{n}}
        {#1}{\mathbb{R}^{#1}}
        % You can add a finite number of more cases :)
   }%
}%

\newcommand{\fun}[3][f]{%
    \IfEqCase{#1}{
        {f}{f\colon\mathcal{A}\rightarrow\mathcal{B}}
        {g}{g\colon\mathcal{C}\rightarrow\mathcal{D}}
        {#1}{#1\colon #2\rightarrow #3}
% Maybe "othercase" can generate case like f or g.
    }
}

\begin{document}

If I use $f$ function like this $\fun[f]{\rn}{\rn[m]}$,

but I expect to use a function called $f\colon\rn\rightarrow\rn[m]$.

This example works fine, e.g, $\fun[\omega]{\rn}{\rn[1]}$ but I want to use 
this command

%\fun[othercase]{f}{\rn}{\rm[1]}

\end{document}

This example works fine, e.g, $\fun[\omega]{\rn}{\rn[1]}$.

If I use $f$ function like this $\fun[o]{f}{\rn}{\rn[m]}$,

but I expect to use a function called $f\colon\rn\rightarrow\rn[m]$.

Update

I tried to construct the command like this:

\newcommand{\fun}[5][default]{%
    \IfEqCase{#1}{
        {default}{
            \IfEqcase{#2}{
                {f}{#2\colon\mathcal{A}\rightarrow\mathcal{B}}
                {g}{#2\colon\mathcal{C}\rightarrow\mathcal{D}}
            }
        }
        {other}{
            {}{#3\colon #4 \rightarrow #5}
        }
    }
}

Because I don't want lost the default options with f (f\colon\mathcal{A}\rightarrow\mathcal{B}) and g (g\colon\mathcal{C}\rightarrow\mathcal{D}) respectively.

I see that I need one optional argument inside other optional argument.

Source

I use this idea for create the command.

• How to make a command with three optional arguments?

Thanks in advance!

Answer 1

It's very unclear to me what you are trying to do here, or why (given the output you expect) you have defined \fun as you do.

Trace the expansion (use \tracingmacros=1 if it helps, though there's a lot of junk from xstring):

• You call \fun[f]{\rn}{\rn[m]}

• After dealing with the optional argument, this boils down to using a command which has the template \fun[#1]{#2}{#3}. In your first example these #1=f, #2=\rn and #3=rn[m]`.

• Within \fun the first thing you do is use a case statement to consider the optional argument #1. The first line matches if #1 is f, which it is.

• That line is therefore expanded, to \mathcal{A}\rightarrow\mathcal{B}

• Since you haven't asked \fun to do anything with #2 and #3 in such a case, they just disappear, like any other unused argument.

What I don't understand is why you have this definition of \fun if you want what you say you want: I can't understand why you are looking at the first optional argument at all. Looks to me like you just want:

\newcommand{\fun}[3]{%
  \ensuremath{#1:#2#3}}

And then use it as \fun{f}{\rn}{\rn[m]}

Answer 2

If you uncomment the line you say you want you get an error that there is missing $ as you have omitted the $ if you surround the expression with $ you get the error ! Class memoir Error: Font command \rm is not supported. because \rm is not defined to do anything. perhaps you intended \rn but neither of those errors are comnnected with your macros at all.

I do not see why you are using the test in \fun where it is not doing anything useful (in the first two cases it means you discard the arguments??) the xstring test isn't doing much in \rn either (only the test for 1, omitting the superscript, is doing anything useful) but I left that.

\documentclass{memoir}
\usepackage{amsmath,amssymb}
\usepackage{xstring}
\newcommand{\rn}[1][n]{%
    \IfEqCase{#1}{%
        {1}{\mathbb{R}}%
        {2}{\mathbb{R}^{2}}%
        {3}{\mathbb{R}^{3}}%
        {n}{\mathbb{R}^{n}}
        {#1}{\mathbb{R}^{#1}}
        % You can add a finite number of more cases :)
   }%
}%

\newcommand{\fun}[3][f]{%
        #1\colon #2\rightarrow #3
}

\begin{document}

If I use $f$ function like this $\fun[f]{\rn}{\rn[m]}$,

but I expect to use a function called $f\colon\rn\rightarrow\rn[m]$.

This example works fine, e.g, $\fun[\omega]{\rn}{\rn[1]}$ but I want to use 
this command

$\fun[f]{\rn}{\rn[1]}$

\end{document} 

Answer 3

In this small example, I created two subcommands, the first one I called default and the second general, I used the xstring package the command \IfEqCase, but I have obtained some unexpected results, I hope it helps you.

\documentclass[14pt]{memoir}
\usepackage{amsmath,amsfonts,amssymb}
\usepackage{xstring}

\newcommand{\comdefault}[1][f]{
    \IfEqCase{#1}{
        {f}{\ensuremath{f\colon\mathcal{A}\rightarrow\mathcal{B}}}
        {g}{\ensuremath{g\colon\mathcal{C}\rightarrow\mathcal{D}}}}%[other case]
}

\newcommand{\comgeneral}[3]{
    \ensuremath{#1\colon#2\rightarrow#3}
}

\newcommand{\comfinal}[4][default]{
    \IfEqCase{#1}{
        {default}{\comdefault[#2]}
        {general}{\comgeneral{#2}{#3}{#4}}}
}

\begin{document}

\begin{enumerate}
    \item The function $g$ is \comdefault[g]. % Subcommand 1
    \item The function $f$ is \comgeneral{f}{X}{Y}. % Subcommand 2
    \item La función $h_1$ does not work correctly %\comfinal{h_1}. %{}%{}{}
    \item La función $h_2$ does not work correctly either. %\comfinal{h_2}{}
    \item The function $h_3$ works \comfinal[general]{X}{Y}{Z}
\end{enumerate}
\end{document}