Using loop in newcommand to build longer equal sign

Question

Following Is there a wider equal sign? and "For loop" in newcommand, I have come up with the naive command:

\newcount\tmp
\newcommand{\eqlong}[1][2]{% need this to prevent extra vertical space
    =
    \tmp=0
    \loop
        % increment dummy counter
        \advance\tmp by 1
        \joinrel=
        % repeat the loop provided the counter is within specified bound
        \ifnum\tmp<#1
    \repeat
}

Supposedly, \eqlong3 for instance should produce =\joinrel=\joinrel=, i.e. an equal sign about 3 times as long as a normal equal sign. The command as I've written doesn't work, but I'm wondering if someone could fix it.

I know a post I linked above (and also Extendible equals sign) mention the extarrows package, but here I just want a symbol whose length I can control with a parameter I pass.

Your command defines an optional argument delimited by square brackets, so the syntax should be \eqlong[3] (or \eqlong with no argument). See this tutorial on the LaTeX syntax of \newcommand. — mbert, Oct 31 '22 at 06:42
As to the actual command, note that \eqlong[3] actually defines a symbol that is four times as long as = (think about where in the code the loop checks whether or not to terminate). — mbert, Oct 31 '22 at 06:48
% need this to prevent extra vertical space can not be right. It does save one token of memory (as would % after the two =) — David Carlisle, Oct 31 '22 at 10:28

score 5 · Answer 1 · answered Oct 31 '22 at 09:05

You can make the command as a one-liner:

\documentclass{article}
\ExplSyntaxOn
\NewDocumentCommand{\eqlong}{O{2}}
 {
  =\prg_replicate:nn { #1 - 1 } { \joinrel= }
 }
\ExplSyntaxOff
\begin{document}
$a=b$
$a\eqlong[1] b$% the same as before
$a\eqlong b$% default, twice as long
$a\eqlong[3] b$% three times as long
\end{document}

This of course assumes that the argument you pass is a positive integer. If you pass something else, you'll get an error.

Now let's try and solve a more complicated problem: we want that \eqlong[3] produces a relation symbol exactly three times long as the equal sign.

\documentclass{article}
\makeatletter
\NewDocumentCommand{\eqlong}{O{2}}{%
  \mathrel{\mathpalette\eqlong@{#1}}%
}
\NewDocumentCommand{\eqlong@}{mm}{%
  \begingroup
  \sbox\z@{$\m@th#1=$}%
  \ifnum#2>1
    \makebox[#2\wd\z@][s]{%
      \copy\z@
      \kern-0.5\wd\z@
      \cleaders\hbox{\kern-0.4\wd\z@\copy\z@\kern-0.4\wd\z@}\hfill
      \kern-0.5\wd\z@
      $\m@th#1\mkern-8mu$
      \copy\z@
    }%
  \else
    \copy\z@
  \fi
  \endgroup
}
\makeatother
\begin{document}
$a=b$
$a\eqlong[1] b$% the same as before
$a\eqlong b$% default, twice as long
$a\eqlong[3] b$% three times as long
$a\eqlong[9] b$
\sbox0{${=}{=}{=}{=}{=}$}\the\wd0
\sbox0{$\eqlong[5]$}\the\wd0
$\scriptstyle a=b$
$\scriptstyle a\eqlong[1] b$
$\scriptstyle a\eqlong b$
$\scriptstyle a\eqlong[2]b$
\end{document}

As you see, replicating five equals signs has the same width as \eqlong[5].

score 4 · Accepted Answer · answered Oct 31 '22 at 07:21

Your command defines an optional argument that, if given, must be delimited by square brackets. If I understand your request correctly, you also want \eqlong[n] to produce a symbol n-times the length of =. As given, it produces a symbol (n+1)-times the length of =. This is fixed by simply changing \tmp=0 to \tmp=1.

\documentclass{article}
\newcount\tmp
\newcommand{\eqlong}[1][2]{% need this to prevent extra vertical space
    =
    \tmp=1
    \loop
        % increment dummy counter
        \advance\tmp by 1
        \joinrel=
        % repeat the loop provided the counter is within specified bound
        \ifnum\tmp<#1
    \repeat
}
\begin{document}
$a=b$
$a\eqlong b$ %default, twice as long
$a\eqlong[3] b$ %three times as long
\end{document}

ahh, I see I messed that up. I thought there was a deeper issue. My bad. — D.R, Oct 31 '22 at 07:49

score 4 · Answer 3 · answered Oct 31 '22 at 12:35

4

In OpTeX, you can make the command as a one-liner:

\optdef\eqlong [2]{=\fornum 2..\the\opt \do{\joinrel=}}
$a=b$
$a\eqlong[1] b$% the same as before
$a\eqlong b$% default, twice as long
$a\eqlong[3] b$% three times as long
\bye

answered Oct 31 '22 at 12:35

wipet

74,238

Ulrich Diez · Answer 4 · 2022-10-31T14:06:49.557

\documentclass{article}

\newcount\tmp

\newcommand{\eqlong}[1][2]{%
    \tmp=0
    \loop
    \ifnum\tmp<\numexpr(#1)\relax
      \ifnum\tmp>0 \joinrel\fi
      \advance\tmp by 1
      =
    \repeat
}

\begin{document}

$a\eqlong[0] b$

$a=b$

$a\eqlong[1] b$

$a\eqlong[2] b$

$a\eqlong b$ %default=2

$a\eqlong[3] b$

\end{document}

\documentclass{article}
\newcommand{\eqlong}[1][2]{%
  \ifnum#1>0 %
    =\ifnum#1>1 \joinrel\fi
    \expandafter\eqlong\expandafter[\the\numexpr(#1)-1\expandafter\relax\expandafter]%
  \fi
}
\begin{document}
$a\eqlong[0] b$
$a=b$
$a\eqlong[1] b$
$a\eqlong[2] b$
$a\eqlong b$ %default=2
$a\eqlong[3] b$
\end{document}

Using loop in newcommand to build longer equal sign

4 Answers4