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Can anyone explain me the fine details of the \strut macro. Why is the box \copyied in mathmode but \unhcopyied otherwise?

% latex.ltx, line 484, but identical in plainTeX

The TeXBook states (e.g. page 120) that the unboxing macros unset any glue in the box's outer level, but there is no glue in the \strutbox, is it?

I understand that \strutbox is an actually empty box with a font size dependent height (.7\baselineskip, where \baselineskip is about 1.2x font size) and depth (.3\baselineskip) but zero width:

% latex.ltx, line 2357, inside the `\set@fontsize` macro

I assume it is more efficient to use a box instead of repeatably use \vrule. Also you can use \ht\strutbox and \dp\strutbox to access the height and depth as dimensions. This allows you to write .5\ht\strutbox while .5*.7\baselineskip would require eTeX's \dimexpr or a temporal assignment.

I really like to understand the fine nuances of TeX better.

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up vote 22 down vote accepted

You can't \unhcopy an hbox in math mode, that's simple! Indeed



! Incompatible list can't be unboxed.
<*> $\unhcopy\strutbox

The command \unhcopy is used outside of math mode because of two facts:

  1. it starts horizontal mode;
  2. it removes one level of boxing, avoiding "too deep" nesting.

A similar definition is that of \leavevmode:


where \voidb@x is a box register that should always remain void. The difference between \unhbox and \unhcopy is that the former destroys the contents of the box register, while the latter produces only a copy of it and leaves the register intact.

Among the duties of \fontsize is to provide a convenient \strut; the strut's height and depth are always 7/10 and 3/10 of the baselineskip, which is set by the second argument to \fontsize. So when a command such as \large is executed, also the strut is changed; it will restored when the group ends or another \fontsize command (possibly implicit) is executed.

(For the rest you're right.)


The register \voidb@x can be used to ensure that a box register is void, by saying \setbox\mybox=\box\voidb@x, which destroys the contents of \mybox and is more efficient than the equivalent {\setbox0=\box\mybox}.

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I see, both things make perfect sense: Unboxing a non-math hbox in mathmode wouldn't be right and removing one unnecessary boxing level is of course more efficient. Nice answer! (As usual, I'm waiting a little bit before accepting it) – Martin Scharrer Jun 28 '11 at 18:07

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