Error with a command whose purpose is to allow a value to adapt over the course of a document

Question

I am trying to write a command that will do the following, in order:

Measure the width of an input word.
Compare this value to an existing value.
If the input word's width is greater, then take on that measure as value. Otherwise, do nothing.

What is going wrong with the following code?

\documentclass{article}
\usepackage{ifthen}
\usepackage{calc}
\usepackage{showframe}
\newlength\shape
\setlength{\shape}{1mm}
\newcommand{\shifter}[1]{\ifthenelse
{\shape<\widthof{#1}}              % If the width of the input word is greater than the length of \shape,
{\setlength{\shape}{\widthof{#1}}} % then make the length of \shape equal to the width of the input word.
{}                                 % Otherwise, do nothing.
}%
\setlength{\parindent}{0pt}
\begin{document}
\hspace{\shape}Hello % Output as expected: `Hello' is indented 1mm from the left margin.
\shifter{Tyrannosaurus}
\hspace{\shape}Hello % Output not as expected. I wanted Hello' to be indented the width of the wordTyrannosaurus' from the left margin. Instead, `Hello' is indented 1mm from the left margin.
\end{document}

Possibly relevant: I am getting the following error:

A number should have been here; I inserted `0'.

Why?

Your question sounds a bit like an x-y-problem. Can you add a bit more background what you are actually trying to do? — samcarter_is_at_topanswers.xyz, Aug 21 '23 at 20:06
\widthof{#1} is not a length that can be used in ifthenelse, it is a special construct just valid in calc package expressions (which includes \setlength) — David Carlisle, Aug 21 '23 at 21:54
@samcarter_is_at_topanswers.xyz Ultimately, I am trying to somewhat alter Zarko's answer (https://tex.stackexchange.com/a/693984/277990) to a previous question I had so that it works without obliging the writer to manually plug in the widest word of a group of words (and find out which one is the widest to begin with). See \settowidth\colwidth{Yessirino} % <--- select the longest text in the last columns in Zarko's code in the above link. (The purpose of \colwidth is to locally establish a column width.) — Noah J, Aug 21 '23 at 23:19

lukeflo · Accepted Answer · 2023-08-22T08:49:05.030

1

You neither need ifthen nor calc. Just define a second dimen for comparison, set it to the width with settowidth (really descriptive) and use a more robust condition of etoolbox:

\documentclass{article}
\usepackage{etoolbox}
\usepackage{showframe}
\newdimen\shape
\setlength{\shape}{1mm}
\newdimen\shapecompare
\newcommand{\shifter}[1]{%
\settowidth\shapecompare{#1}%
\ifdimless{\shape}{\shapecompare}%
{\setlength{\shape}{\shapecompare}}%
{}%
}%
\setlength{\parindent}{0pt}
\begin{document}
\hspace{\shape}Hello
\shifter{Tyrannosaurus}
Tyrannosaurus
\hspace{\shape}Hello
\end{document}

edited Aug 22 '23 at 08:49

answered Aug 21 '23 at 19:33

lukeflo

1,555

1

! Misplaced alignment tab character &. – David Carlisle Aug 21 '23 at 21:51
Oh, just a typo, should have been a %. Wrote ot on my smartphone. Thanks for the hint ;) – lukeflo Aug 22 '23 at 04:55
From where do you deduce that \ifthenelse is deprecated? – egreg Aug 22 '23 at 08:26
@egreg thought ifthen is not recommended, but I may be wrong. Edited in the post. – lukeflo Aug 22 '23 at 08:46
My suggestion was related to opinions like mentioned here, but even if the package may have some drawbacks it is, of course, not deprecated or obsolete. – lukeflo Aug 22 '23 at 09:01

score 0 · Answer 2 · answered Aug 22 '23 at 08:25

There are a few problems with your code.

First problem: if you want to compare lengths with \ifthenelse you need \lengthtest (only comparison between integers needs nothing).

Second problem: \widthof doesn't work in \lengthtest.

Fix: use a scratch dimension register.

\documentclass{article}
\usepackage{ifthen}
\usepackage{showframe}
\newlength\shape
\setlength{\shape}{1mm}
\newcommand{\shifter}[1]{%
  \settowidth{\dimen0}{#1}%
  \ifthenelse{\lengthtest{\shape<\dimen0}}{\setlength{\shape}{\dimen0}}{}%
}
\setlength{\parindent}{0pt}
\begin{document}
\hspace{\shape}Hello
\shifter{Tyrannosaurus}
\hspace{\shape}Hello
Tyrannosaurus % check the space
\end{document}

You don't need ifthen, actually:

\newcommand{\shifter}[1]{%
  \settowidth{\dimen0}{#1}%
  \ifdim\shape<\dimen0
    \setlength{\shape}{\dimen0}%
  \fi
}

Error with a command whose purpose is to allow a value to adapt over the course of a document

2 Answers2