[dcl.array,expr.sub] Consolidate notes on symmetry of array subscripting.

Author: jensmaurer

source/declarators.tex

@@ -1183,31 +1183,8 @@
 
 \pnum
 \begin{note}
-Except where it has been declared for a class\iref{over.sub},
-the subscript operator
-\tcode{[]}
-is interpreted
-in such a way that
-\tcode{E1[E2]}
-is identical to
-\tcode{*((E1)+(E2))}\iref{expr.sub}.
-Because of the conversion rules
-that apply to
-\tcode{+},
-if
-\tcode{E1}
-is an array and
-\tcode{E2}
-an integer,
-then
-\tcode{E1[E2]}
-refers to the
-\tcode{E2}-th
-member of
-\tcode{E1}.
-Therefore,
-despite its asymmetric
-appearance, subscripting is a commutative operation.
+The subscript operator can be overloaded for a class\iref{over.sub}.
+For the operator's built-in meaning, see \ref{expr.sub}.
 \end{note}
 
 \rSec2[dcl.fct]{Functions}%

source/expressions.tex

@@ -2145,14 +2145,19 @@
 is true even if the subscript operator is used in the following common idiom:
 \tcode{\&x[0]}.}
 The expression \tcode{E1[E2]} is identical (by definition) to
-\tcode{*((E1)+(E2))}
-\begin{note}
-see~\ref{expr.unary} and~\ref{expr.add} for details of \tcode{*} and
-\tcode{+} and~\ref{dcl.array} for details of arrays.
-\end{note}, except that in the case of an array operand, the result is an lvalue
+\tcode{*((E1)+(E2))},
+except that in the case of an array operand, the result is an lvalue
 if that operand is an lvalue and an xvalue otherwise.
 The expression \tcode{E1} is sequenced before the expression \tcode{E2}.
 
+\pnum
+\begin{note}
+Despite its asymmetric appearance, subscripting is a commutative
+operation except for sequencing.
+See~\ref{expr.unary} and~\ref{expr.add} for details of \tcode{*} and
+\tcode{+} and~\ref{dcl.array} for details of array types.
+\end{note}
+
 \pnum
 A \grammarterm{braced-init-list} shall not be used with the built-in subscript operator.