\begin{codeblock}

a = (a + (b + 32765));

\end{codeblock}

respectively, 4 and $-8$ or $-17$ and 12. However on a machine in which

overflows do not produce an exception and in which the results of

overflows are reversible, the above expression statement can be

\begin{note}

Historically, lvalues and rvalues were so-called

(although this is no longer generally true);

glvalues are ``generalized'' lvalues,

prvalues are ``pure'' rvalues,

\begin{note}

There are no prvalue bit-fields; if a bit-field is converted to a

prvalue\iref{conv.lval}, a prvalue of the type of the bit-field is

\end{note}

can be converted to type \tcode{T2}

if the qualification-combined type of \tcode{T1} and \tcode{T2} is \tcode{T2}.

\begin{note}

type \keyword{const} \tcode{T**} (that is, if line \#1 below were

\begin{codeblock}

int main() {

const char c = 'c';

\begin{note}

representation different from the original value.

\end{note}

and the converted operand is used in place of the original operand for

the remainder of this subclause.

\begin{note}

\end{note}