+\begin{figure}
+ \centering
+ \begin{tikzpicture}[x=7.5mm, y=-14mm, baseline=(current bounding box.east)]
+ \node[lit] at ( 0, 0) (R) {SodObject};
+ \node[lit] at (-3, +1) (A) {A}; \draw[->] (A) -- (R);
+ \node[lit] at (-1, +1) (B) {B}; \draw[->] (B) -- (R);
+ \node[lit] at (+1, +1) (C) {C}; \draw[->] (C) -- (R);
+ \node[lit] at (+3, +1) (D) {D}; \draw[->] (D) -- (R);
+ \node[lit] at (-2, +2) (E) {E}; \draw[->] (E) -- (A);
+ \draw[->] (E) -- (B);
+ \node[lit] at (+2, +2) (F) {F}; \draw[->] (F) -- (A);
+ \draw[->] (F) -- (D);
+ \node[lit] at (-1, +3) (G) {G}; \draw[->] (G) -- (E);
+ \draw[->] (G) -- (C);
+ \node[lit] at (+1, +3) (H) {H}; \draw[->] (H) -- (F);
+ \node[lit] at ( 0, +4) (I) {I}; \draw[->] (I) -- (G);
+ \draw[->] (I) -- (H);
+ \end{tikzpicture}
+ \quad
+ \vrule
+ \quad
+ \begin{minipage}[c]{0.45\hsize}
+ \begin{nprog}
+ class A: SodObject \{ \}\quad\=@/* @|A|, @|SodObject| */ \\
+ class B: SodObject \{ \}\>@/* @|B|, @|SodObject| */ \\
+ class C: SodObject \{ \}\>@/* @|B|, @|SodObject| */ \\
+ class D: SodObject \{ \}\>@/* @|B|, @|SodObject| */ \\+
+ class E: A, B \{ \}\quad\=@/* @|E|, @|A|, @|B|, \dots */ \\
+ class F: A, D \{ \}\>@/* @|F|, @|A|, @|D|, \dots */ \\+
+ class G: E, C \{ \}\>@/* @|G|, @|E|, @|A|,
+ @|B|, @|C|, \dots */ \\
+ class H: F \{ \}\>@/* @|H|, @|F|, @|A|, @|D|, \dots */ \\+
+ class I: G, H \{ \}\>@/* @|I|, @|G|, @|E|, @|H|, @|F|,
+ @|A|, @|B|, @|C|, @|D|, \dots */
+ \end{nprog}
+ \end{minipage}
+
+ \caption{An example class graph and class precedence lists}
+ \label{fig:concepts.classes.cpl-example}
+\end{figure}
+
+\begin{example}
+ Consider the class relationships shown in
+ \xref{fig:concepts.classes.cpl-example}.
+
+ \begin{itemize}
+
+ \item @|SodObject| has no proper superclasses. Its class precedence list
+ is therefore simply $\langle @|SodObject| \rangle$.
+
+ \item In general, if $X$ is a direct subclass only of $Y$, and $Y$'s class
+ precedence list is $\langle Y, \ldots \rangle$, then $X$'s class
+ precedence list is $\langle X, Y, \ldots \rangle$. This explains $A$,
+ $B$, $C$, $D$, and $H$.
+
+ \item $E$'s list is found by merging its local precedence list $\langle E,
+ A, B \rangle$ with the class precedence lists of its direct superclasses,
+ which are $\langle A, @|SodObject| \rangle$ and $\langle B, @|SodObject|
+ \rangle$. Clearly, @|SodObject| must be last, and $E$'s local precedence
+ list orders the rest, giving $\langle E, A, B, @|SodObject|, \rangle$.
+ $F$ is similar.
+
+ \item We determine $G$'s class precedence list by merging the three lists
+ $\langle G, E, C \rangle$, $\langle E, A, B, @|SodObject| \rangle$, and
+ $\langle C, @|SodObject| \rangle$. The class precedence list begins
+ $\langle G, E, \ldots \rangle$, but the individual lists don't order $A$
+ and $C$. Comparing these to $G$'s direct superclasses, we see that $A$
+ is a subclass of $E$, while $C$ is a subclass of -- indeed equal to --
+ $C$; so $A$ must precede $C$, as must $B$, and the final list is $\langle
+ G, E, A, B, C, @|SodObject| \rangle$.
+
+ \item Finally, we determine $I$'s class precedence list by merging $\langle
+ I, G, H \rangle$, $\langle G, E, A, B, C, @|SodObject| \rangle$, and
+ $\langle H, F, A, D, @|SodObject| \rangle$. The list begins $\langle I,
+ G, \ldots \rangle$, and then we must break a tie between $E$ and $H$; but
+ $E$ is a subclass of $G$, so $E$ wins. Next, $H$ and $F$ must precede
+ $A$, since these are ordered by $H$'s class precedence list. Then $B$
+ and $C$ precede $D$, since the former are superclasses of $G$, and the
+ final list is $\langle I, G, E, H, F, A, B, C, D, @|SodObject| \rangle$.
+
+ \end{itemize}
+
+ (This example combines elements from \cite{Barrett:1996:MSL} and
+ \cite{Ducournau:1994:PMM}.)
+\end{example}
+