Savebox and usebox unexpected behaviour for the box with index 10

Question

I have found that when saving and later using boxes the box with index 10 behaves differently to all other.

Here is simple demo that shows different behaviour. In preamble I define this commands:

\usepackage{forloop}
\newcounter{BoxCount}
\newcommand{\numberedBox}[1]{%
  \stepcounter{BoxCount}%
  \savebox{\theBoxCount}{\vbox{\noindent\theBoxCount\\~ #1}}%
}%
\newcommand{\writeBoxes}{
  \stepcounter{BoxCount}%
  \newcounter{I}%
  \setcounter{I}{1}%
  \forloop{I}{1}{\value{I} < \value{BoxCount}}{%
  \usebox{\theI}\par
}%

The \numbredBox basically stores given text in indexed box while adding number of the box to the text. The \writeBoxes prints all stored boxes in order they were stored.

In document body I have this code:

\numberedBox{word}
\numberedBox{word}
\numberedBox{word}
\numberedBox{word}
\numberedBox{word}
\numberedBox{word}
\numberedBox{word}
\numberedBox{word}
\numberedBox{word}
\numberedBox{word}
\numberedBox{word}
\numberedBox{word}
\numberedBox{word}
\writeBoxes

This is printed output of this document: All the boxes are printed as expected except for 10th box which somehow interfere with box 11. I have played with similar code a little and found out that 10th saved almost always box behaves strange. In some cases I even got ! Incompatible list can't be unboxed error but only for 10th saved box (content of all the boxes was equal).

Why is this happening? How can I avoid it?

---

EDIT

Here is complete source of this example that Hebert asked for. Just compile with pdfLaTeX and you will get output shown above.

\documentclass[a5paper]{book}
\usepackage{forloop}
\newcounter{BoxCount}
\newcommand{\numberedBox}[1]{%
  \stepcounter{BoxCount}%
  \savebox{\theBoxCount}{\vbox{\noindent\theBoxCount\\~ #1}}%
}%
\newcommand{\writeBoxes}{
  \stepcounter{BoxCount}%
  \newcounter{I}%
  \setcounter{I}{1}%
  \forloop{I}{1}{\value{I} < \value{BoxCount}}{%
    \usebox{\theI}\par
  }%
}
\begin{document}
  \numberedBox{word}
  \numberedBox{word}
  \numberedBox{word}
  \numberedBox{word}
  \numberedBox{word}
  \numberedBox{word}
  \numberedBox{word}
  \numberedBox{word}
  \numberedBox{word}
  \numberedBox{word}
  \numberedBox{word}
  \numberedBox{word}
  \writeBoxes
\end{document}

Answer 1

use the LaTeX definition of boxes:

\documentclass[a5paper]{book}
\usepackage{forloop}
\newcounter{BoxCount}
\setcounter{BoxCount}{0}
\newcounter{I}
\newcommand\numberedBox[1]{%
  \stepcounter{BoxCount}%
  \expandafter\newsavebox\expandafter{\csname BOX\alph{BoxCount}\endcsname}
  \expandafter\sbox\expandafter{\csname BOX\alph{BoxCount}\endcsname}{\vbox{\noindent\theBoxCount\\~ #1}}%
}%
\newcommand{\writeBoxes}{
  \stepcounter{BoxCount}%
  \setcounter{I}{1}%
  \forloop{I}{1}{\value{I} < \value{BoxCount}}{%
    \usebox{\csname BOX\alph{I}\endcsname}\par
  }%
}
\begin{document}
  \numberedBox{word}
  \numberedBox{word}
  \numberedBox{word}
  \numberedBox{word}
  \numberedBox{word}
  \numberedBox{word}
  \numberedBox{word}
  \numberedBox{word}
  \numberedBox{word}
  \numberedBox{word}
  \numberedBox{word}
  \numberedBox{word}
  \writeBoxes
\end{document}

Answer 2

You shouldn't be using box registers by number other than those numbered from 0 to 9.

In particular box register 10 corresponds to \voidb@x which the LaTeX kernel considers as an always empty box. If you fill it with something a number of macros will work wrongly. Among them

\leavevmode
\settowidth \settoheight \settowidth

Why do you get an error? Because \usebox does \leavevmode which in turn does \unhbox\voidb@x; you've set \box10 (that is \voidb@x) to a \vbox, so the command performed by \leavevmode tries to \unhbox a \vbox, which is not allowed.

Here's a safer (but not completely foolproof) way. If it's just for experiments, then it's not so risky; on the other hand I would never use this for real document production. In that case I would allocate my box registers.

\documentclass[a5paper]{book}
\usepackage{forloop,etex}
\newcounter{BoxCount}
\setcounter{BoxCount}{\count14}
\ifnum\value{BoxCount}<256 
  \setcounter{BoxCount}{355}
\else
  \addtocounter{BoxCount}{100}
\fi
\mathchardef\FirstBoxCount=\value{BoxCount}
\newcommand{\numberedBox}[1]{%
  \stepcounter{BoxCount}%
  \savebox{\theBoxCount}{\vbox{\noindent\theBoxCount\\~ #1}}%
}
\newcounter{I}
\newcommand{\writeBoxes}{%
  \stepcounter{BoxCount}%
  \setcounter{I}{\FirstBoxCount}%
  \forloop{I}{1}{\value{I} < \value{BoxCount}}{%
    \usebox{\theI}\par
  }%
}
\begin{document}
  \numberedBox{word}
  \numberedBox{word}
  \numberedBox{word}
  \numberedBox{word}
  \numberedBox{word}
  \numberedBox{word}
  \numberedBox{word}
  \numberedBox{word}
  \numberedBox{word}
  \numberedBox{word}
  \numberedBox{word}
  \numberedBox{word}
  \writeBoxes
\end{document}

Using the extended register set and going 100 registers from the last allocated one if the extended register set is already used, this should work without touching any allocated box register.

---

The LaTeX kernel allocates more box than register 10:

\voidb@x=\box10
\strutbox=\box11
\@tempboxa=\box12
\rootbox=\box13
\@labels=\box14
\@curline=\box15
\@curfield=\box16
\@tabfbox=\box17
\@arstrutbox=\box18
\@picbox=\box19
\@linechar=\box20
\@dashbox=\box21
\@begindvibox=\box22
\@outputbox=\box23
\@leftcolumn=\box24
\@holdpg=\box25

Using these registers and following ones possibly allocated by packages is an almost sure way to make things go wrong. You're shooting your own foot if you do it.

---

Here is a completely safe version.

\documentclass[a5paper]{book}
\usepackage{forloop,etex}
\newcounter{BoxCount}

\makeatletter
\newcommand{\myBox}[1]{\@nameuse{myBox@\number\value{#1}}}
\newcommand{\checkBox}[1]{%
  \@ifundefined{myBox@\number\value{#1}}
    {\expandafter\newbox\csname myBox@\number\value{#1}\endcsname}
    {}%
}
\makeatother

\newcommand{\numberedBox}[1]{%
  \stepcounter{BoxCount}%
  \checkBox{BoxCount}%
  \savebox{\myBox{BoxCount}}{\vbox{\noindent\theBoxCount\\~ #1}}%
}
\newcounter{I}
\newcommand{\writeBoxes}{%
  \stepcounter{BoxCount}%
  \setcounter{I}{1}%
  \forloop{I}{1}{\value{I} < \value{BoxCount}}{%
    \usebox{\myBox{I}}\par
  }%
}
\begin{document}
  \numberedBox{word}
  \numberedBox{word}
  \numberedBox{word}
  \numberedBox{word}
  \numberedBox{word}
  \numberedBox{word}
  \numberedBox{word}
  \numberedBox{word}
  \numberedBox{word}
  \numberedBox{word}
  \numberedBox{word}
  \numberedBox{word}
  \writeBoxes
\end{document}

When you set a box associated to a counter's value, you first check for its allocation:

\checkBox{BoxCount}

will allocate a new box register if you haven't done it so far; if the box register is already allocated, nothing happens. Then you use

\myBox{<counter>}

when you need to save or use the box register corresponding to the counter's value.