In the memoir manual speaks about

measuring the length of the lowercase alphabet

in order to decide the numbers of characters per line. Can this be done with LaTeX?

  • Both answers seem to be equal correct and elegant (at least to my eyes). I can't decide which one I should accept... – pmav99 May 4 '11 at 19:21
  • My answer is the "official" LaTeX way, while Yiannis one uses plainTeX macros. I personally actually use that macros more often because you only need to box the content once and get the height, width and depth using \ht, \wd and \dp respectively. – Martin Scharrer May 4 '11 at 20:07

If with the length it means the width of all lowercase letters than you can use \settowidth{<length register>}{<content>}:


(I took the liberty to use the same format as Yiannis answer to allow for better comparison. His answer is fine too, but I wanted to show the official LaTeX way.)


You can place the letters in a box and measure its length. Here is a minimal to do this.

  • @pmav99 \textgreek{parakal`w} – Yiannis Lazarides May 4 '11 at 19:32
  • @Yiannis You are hardcore! This is the Unicode age :P – pmav99 May 4 '11 at 19:47
  • I finally chose Martin's answer as he uses one less `\` ... LOL – pmav99 May 4 '11 at 19:49
  • 2
    @pmav99 I see, my solution was optimized for the minimum number of curly brackets! – Yiannis Lazarides May 5 '11 at 3:33
  • @Yiannis: then replace \sbox\alphabet{...} by \savebox\alphabet\hbox\bgroup...\egroup. – Bruno Le Floch May 14 '11 at 13:16

Just for the record, page 15 of memman.pdf brings the following to print the length of the lowercase alphabet:

\newlength{\mylen}                % a length
\newcommand{\alphabet}{abc...xyz} % the lowercase alphabet
\begingroup                       % keep font change local
% font specification e.g., \Large\sffamily
The length of this alphabet is \the\mylen. % print in document
\typeout{The length of the Large sans alphabet
    is \the\mylen}                         % put in log file
\endgroup                         % end the grouping

Which is basically @Martin's way.

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