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[meta.logical] use math mode subscripts for Bi, B1, BN etc. #2549

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Dec 3, 2018
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[meta.logical] use math mode subscripts for Bi, B1, BN etc. #2549

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38 changes: 20 additions & 18 deletions source/utilities.tex
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Original file line number Diff line number Diff line change
Expand Up @@ -18337,32 +18337,33 @@
forms the logical conjunction of its template type arguments.

\pnum
For a specialization \tcode{conjunction<B1, ..., BN>},
if there is a template type argument \tcode{Bi} for which \tcode{bool(Bi::value)} is \tcode{false},
then instantiating \tcode{conjunction<B1, ..., BN>::value}
does not require the instantiation of \tcode{Bj::value} for \tcode{j > i}.
For a specialization \tcode{conjunction<$\tcode{B}_{1}$, $\dotsc$, $\tcode{B}_{N}$>},
if there is a template type argument $\tcode{B}_{i}$
for which \tcode{bool($\tcode{B}_{i}$::value)} is \tcode{false},
then instantiating \tcode{conjunction<$\tcode{B}_{1}$, $\dotsc$, $\tcode{B}_{N}$>::value}
does not require the instantiation of \tcode{$\tcode{B}_{j}$::value} for $j > i$.
\begin{note} This is analogous to the short-circuiting behavior of
the built-in operator \tcode{\&\&}.
\end{note}

\pnum
Every template type argument
for which \tcode{Bi::value} is instantiated
for which \tcode{$\tcode{B}_{i}$::value} is instantiated
shall be usable as a base class and
shall have a member \tcode{value} which
is convertible to \tcode{bool},
is not hidden, and
is unambiguously available in the type.

\pnum
The specialization \tcode{conjunction<B1, ..., BN>}
The specialization \tcode{conjunction<$\tcode{B}_{1}$, $\dotsc$, $\tcode{B}_{N}$>}
has a public and unambiguous base that is either
\begin{itemize}
\item
the first type \tcode{Bi} in the list \tcode{true_type, B1, ..., BN}
for which \tcode{bool(Bi::value)} is \tcode{false}, or
the first type $\tcode{B}_{i}$ in the list \tcode{true_type, $\tcode{B}_{1}$, $\dotsc$, $\tcode{B}_{N}$}
for which \tcode{bool($\tcode{B}_{i}$::value)} is \tcode{false}, or
\item
if there is no such \tcode{Bi}, the last type in the list.
if there is no such $\tcode{B}_{i}$, the last type in the list.
\end{itemize}
\begin{note} This means a specialization of \tcode{conjunction}
does not necessarily inherit from
Expand All @@ -18386,30 +18387,31 @@
forms the logical disjunction of its template type arguments.

\pnum
For a specialization \tcode{disjunction<B1, ..., BN>},
if there is a template type argument \tcode{Bi} for which \tcode{bool(Bi::value)} is \tcode{true},
then instantiating \tcode{disjunction<B1, ..., BN>::value}
does not require the instantiation of \tcode{Bj::value} for \tcode{j > i}.
For a specialization \tcode{disjunction<$\tcode{B}_{1}$, $\dotsc$, $\tcode{B}_{N}$>},
if there is a template type argument $\tcode{B}_{i}$
for which \tcode{bool($\tcode{B}_{i}$::value)} is \tcode{true},
then instantiating \tcode{disjunction<$\tcode{B}_{1}$, $\dotsc$, $\tcode{B}_{N}$>::value}
does not require the instantiation of \tcode{$\tcode{B}_{j}$::value} for $j > i$.
\begin{note} This is analogous to the short-circuiting behavior of
the built-in operator \tcode{||}.
\end{note}

\pnum
Every template type argument
for which \tcode{Bi::value} is instantiated
for which \tcode{$\tcode{B}_{i}$::value} is instantiated
shall be usable as a base class and
shall have a member \tcode{value} which
is convertible to \tcode{bool},
is not hidden, and
is unambiguously available in the type.

\pnum
The specialization \tcode{disjunction<B1, ..., BN>}
The specialization \tcode{disjunction<$\tcode{B}_{1}$, $\dotsc$, $\tcode{B}_{N}$>}
has a public and unambiguous base that is either
\begin{itemize}
\item the first type \tcode{Bi} in the list \tcode{false_type, B1, ..., BN}
for which \tcode{bool(Bi::value)} is \tcode{true}, or
\item if there is no such \tcode{Bi}, the last type in the list.
\item the first type $\tcode{B}_{i}$ in the list \tcode{false_type, $\tcode{B}_{1}$, $\dotsc$, $\tcode{B}_{N}$}
for which \tcode{bool($\tcode{B}_{i}$::value)} is \tcode{true}, or
\item if there is no such $\tcode{B}_{i}$, the last type in the list.
\end{itemize}
\begin{note} This means a specialization of \tcode{disjunction}
does not necessarily inherit from
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