$\langle C, @|SodObject| \rangle$. 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 $\langle
- G, E, A, B, C, @|SodObject| \rangle$.
+ is a superclass of $E$, while $C$ is a superclass 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
+ $E$ is a superclass 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 $\langle I, G, E, H, F, A, B, C, D, @|SodObject| \rangle$.
~mdw / sod / commitdiff