[basic.def.odr] Clarify the meaning of the set of potential results of

an expression and how that relates to odr-use.

Author: zygoloid

source/basic.tex

@@ -270,10 +270,26 @@
 contains the potential results of the right operand.
 \item Otherwise, the set is empty.
 \end{itemize}
+\enternote
+This set is a (possibly-empty) set of \grammarterm{id-expression}{s},
+each of which is either \tcode{e} or a subexpression of \tcode{e}.
+\enterexample
+In the following example, the set of potential results of the initializer
+of \tcode{n} contains the first \tcode{S::x} subexpression, but not the second
+\tcode{S::x} subexpression.
+\begin{codeblock}
+struct S { static const int x = 0; };
+int f(const int &r);
+int n = b ? (1, S::x)  // \tcode{S::x} is not odr-used here
+          : f(S::x);   // \tcode{S::x} is odr-used here, so
+                       // a definition is required
+\end{codeblock}
+\exitexample
+\exitnote
 
 \pnum
 A variable \tcode{x} whose name appears as a
-potentially-evaluated expression \tcode{ex} is \defn{odr-used} unless
+potentially-evaluated expression \tcode{ex} is \defn{odr-used} by \tcode{ex} unless
 applying the lvalue-to-rvalue conversion (\ref{conv.lval}) to \tcode{x} yields
 a constant expression~(\ref{expr.const}) that does not invoke any non-trivial
 functions