Incomplete iffalse using xstring

Question

I have a macros.tex included in the preamble that creates some useful dumb macros. I want to define a macro for writing fractions that adds a brace if the argument is longer than 1 character, what I have so far is:

\usepackage{xstrings}
\newcommand{\fracdivide}[3]{%
    \StrLen{#2}[\arglen]%
    \ifthenelse{\arglen > 1}
    {%
        \frac{#1\left(#2\right)}{#1#3}
    }%
    {%
        \frac{#1#2}{#1#3}
    }%
}

That produces the following error: !Incomplete \iffalse; all text was ignored after line 16.

line 16 is exactly the line where I have the \StrLen command. If I remove it (for example using 2 instead of \arglen), it compiles correctly.

Why do I get this error? And how do I solve it?

Here's a MWE, it seems related to the \text command of amsmath.

\documentclass{minimal}
\usepackage{mathtools}
\usepackage{xstring}
\usepackage{ifthen}
\newcommand{\vect}[1]{\boldsymbol{\mathbf{#1}}}
\newcommand{\fracdivide}[3]{%
    \StrLen{#2}[\arglen]%
    \ifthenelse{\arglen > 1}
    {%
        \frac{#1\left(#2\right)}{#1#3}
    }%
    {%
        \frac{#1#2}{#1#3}
    }%
}
\newcommand{\partialdd}[2]{\fracdivide{\partial}{#1}{#2}}
\newcommand{\ddt}[1]{\partialdd{#1}{t}}

\begin{document}
$\ddt{f^\text{big}}$
\end{document}

Answer 1

You are better using \noexpandargs:

\documentclass{article}
\usepackage{mathtools}
\usepackage{xstring}
\usepackage{ifthen}
\newcommand{\vect}[1]{\boldsymbol{\mathbf{#1}}}
\newcommand{\fracdivide}[3]{%
  \begingroup\noexpandarg
  \StrLen{#2}[\arglen]%
  \ifthenelse{\arglen > 1}
    {%
     \frac{#1(#2)}{#1#3}%
    }%
    {%
     \frac{#1#2}{#1#3}%
    }%
  \endgroup
}
\newcommand{\partialdd}[2]{\fracdivide{\partial}{#1}{#2}}
\newcommand{\ddt}[1]{\partialdd{#1}{t}}

\begin{document}

\[
\ddt{f^{\mathrm{big}}} + \ddt{\Gamma}
\]

\end{document}

Note: I removed the useless (and harmful) \left and \right.

The problem is that commands such as \mathrm or \text won't survive the full expansion performed by default by \StrLen. With \noexpandargs this is avoided.

However, something like \ddt{\vect{x}} will be parenthesized. I don't think this is a really useful macro.

Probably a macro with a *-version is easier to work with.

\documentclass{article}
\usepackage{mathtools}
\usepackage{xparse}

\NewDocumentCommand{\fracdivide}{smmm}{%
  \frac{#2\IfBooleanTF{#1}{(#3)}{#3}}{#2#4}%
}

\NewDocumentCommand{\partialdd}{smm}{%
  \IfBooleanTF{#1}{\fracdivide*}{\fracdivide}{\partial}{#2}{#3}%
}
\NewDocumentCommand{\ddt}{sm}{%
  \IfBooleanTF{#1}{\partialdd*}{\partialdd}{#2}{t}%
}

\newcommand{\vect}[1]{\boldsymbol{\mathbf{#1}}}

\begin{document}

\[
\ddt*{f^{\mathrm{big}}} + \ddt{f_1}
- \ddt{\vect{x}} + \ddt*{\vect{x}+\vect{y}}
\]

\end{document}

Answer 2

Your attempt at extracting the length of the string requires the argument to expand to a string. However, f^\text{big} does not. We can \detokenize the input:

\documentclass{article}

\usepackage{mathtools,xstring}

\newcommand{\fracdivide}[3]{
    \StrLen{\detokenize{#2}}[\arglen]
    \ifnum\arglen>1
      \frac{#1\left(#2\right)}{#1#3}
    \else
      \frac{#1#2}{#1#3}
    \fi
}
\newcommand{\partialdd}[2]{\fracdivide{\partial}{#1}{#2}}
\newcommand{\ddt}[1]{\partialdd{#1}{t}}

\begin{document}
$\ddt{f^\text{big}}$
\end{document}

Answer 3

You need something like this:

\def\ddt#1{{\partial\maybebraces{#1}\over\partial t}}
\def\maybebraces#1{\maybebracesA#1&}
\def\maybebracesA#1#2&{\ifx&#2&#1\else\left(#1#2\right)\fi}


$$\ddt{f^{\rm big}} + \ddt{\Gamma}$$

\bye

In contrast to the other solutions here we do not mix the problem of construction of the fraction (with "partial" sign and with "t" in denominator) with the problem of "maybebraces". And (of course) no special packages (and no LaTeX) is explicitly needed.