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Add new number operator rules to language specification. (#1093)
* Add new number operator rules to language specification. Co-authored-by: Erik Ernst <[email protected]>
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‎specification/dartLangSpec.tex

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@@ -13820,7 +13820,65 @@ \subsubsection{Ordinary Invocation}
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Otherwise, the static analysis of $i$ is performed
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as specified in Section~\ref{bindingActualsToFormals},
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considering $F$ to be the static type of the function to call,
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and the static type of $i$ is as specified there.
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and the static type of $i$ is as specified there,
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except that invocations of methods named \code{remainder}
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or \code{clamp} on subtypes of \code{num} that are not subtypes of \code{Never}
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have special rules similar to those for additive (\ref{additiveExpressions})
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and multipliative (\ref{multiplicativeExpressions}) operators.
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\LMHash{}%
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Let $i$ be an invocation of the form \code{$e$.remainder($e_2$)}
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and let $C$ be the context type of $i$.
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The context type of $e_2$ is then determined as follows:
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If \SubtypeNE{T}{\code{num}} and not \SubtypeNE{T}{\code{Never}}, then:
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\begin{itemize}
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\item{} If \SubtypeNE{\code{int}}{C} and not \SubtypeNE{\code{num}}{C},
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and \SubtypeNE{T}{\code{int}}
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then the context type of $e_2$ is \code{int}.
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\item{} If \SubtypeNE{\code{double}}{C} and not \SubtypeNE{\code{num}}{C},
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and not \SubtypeNE{T}{\code{double}}
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then the context type of $e_2$ is \code{double}.
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\item{} Otherwise the context type of $e_2$ is \code{num}.
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\end{itemize}
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Let further $S$ be the static type of $e_2$.
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If \SubtypeNE{T}{\code{num}} and not \SubtypeNE{T}{\code{Never}}
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and $S$ is assignable to \code{num},
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then the static type of $i$ is determined as follows:
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\begin{itemize}
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\item{} If \SubtypeNE{T}{\code{double}}
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then the static type of $i$ is $T$.
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\item{} Otherwise, if \SubtypeNE{S}{\code{double}}
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and not \SubtypeNE{S}{\code{Never}},
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then the static type of $i$ is \code{double}.
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\item{} Otherwise, if \SubtypeNE{T}{\code{int}},
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\SubtypeNE{S}{\code{int}} and not \SubtypeNE{S}{\code{Never}},
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then the static type of $i$ is \code{int}.
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\item{} Otherwise the static type of $i$ is \code{num}.
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\end{itemize}
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\LMHash{}%
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Let $i$ be an invocation of the form \code{$e$.clamp($e_2$,\,\,$e_3$)},
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where \SubtypeNE{T}{\code{num}} and not \SubtypeNE{T}{\code{Never}},
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and let $C$ be the context type of $i$.
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The context type of $e_2$ and $e_3$ is then determined as follows:
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\begin{itemize}
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\item{} If \SubtypeNE{T}{\code{int}},
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\SubtypeNE{\code{int}}{S} and not \SubtypeNE{\code{num}}{S},
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then the context type of $e_2$ and $e_3$ is \code{int}.
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\item{} If \SubtypeNE{T}{\code{double}},
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\SubtypeNE{\code{double}}{S} and not \SubtypeNE{\code{num}}{S},
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then the context type of $e_2$ and $e_3$ is \code{double}.
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\item{} Otherwise the context type of $e_2$ and $e_3$ is \code{num}.
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\end{itemize}
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Let further $T_2$ be the static type of $e_2$ and $T_3$ be the static type
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of $e_3$.
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\begin{itemize}
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\item{} If all of $T$, $T_2$ and $T_3$ are subtypes of \code{int}, but
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not subtypes of \code{Never}, then the static type of $i$ is \code{int}.
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\item{} If all of $T$, $T_2$ and $T_3$ are subtypes of \code{double}, but
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not subtypes of \code{Never}, then the static type of $i$ is \code{double}.
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\item{} Otherwise the static type of $i$ is \code{num}.
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\end{itemize}
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\LMHash{}%
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It is a compile-time error to invoke an instance method on a type literal
@@ -15913,19 +15971,43 @@ \subsection{Additive Expressions}
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the method invocation \code{\SUPER.$op$($e_2$)}.
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\LMHash{}%
15916-
The static type of an additive expression is usually determined by
15917-
the signature given in the declaration of the operator used.
15918-
However, invocations of the operators \code{+} and \code{-} of class \code{int}
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The static type of an additive expression is usually determined
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by the signature given in the declaration of the operator used.
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However, invocations of the operators \code{+} and \code{-} of
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class \code{int}, \code{double} and \code{num}
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are treated specially by the typechecker.
15920-
The static type of an expression $e_1 + e_2$
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where $e_1$ has static type \code{int}
15922-
is \code{int} if the static type of $e_2$ is \code{int},
15923-
and \code{double} if the static type of $e_2$ is \code{double}.
15924-
The static type of an expression $e_1 - e_2$
15925-
where $e_1$ has static type \code{int}
15926-
is \code{int} if the static type of $e_2$ is \code{int},
15927-
and \code{double} if the static type of $e_2$ is \code{double}.
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\LMHash{}%
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Let $e$ be an additive expression of the form \code{$e_1$ $op$ $e_2$},
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let $T$ be the static type of $e_1$,
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and let $C$ be the context type of $e$.
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If \SubtypeNE{T}{\code{num}} and not \SubtypeNE{T}{\code{Never}}, then
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the context type of $e_2$ is determined as follows:
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\begin{itemize}
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\item{} If \SubtypeNE{\code{int}}{C} and not \SubtypeNE{\code{num}}{C},
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and \SubtypeNE{T}{\code{int}}
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then the context type of $e_2$ is \code{int}.
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\item{} If \SubtypeNE{\code{double}}{C} and not \SubtypeNE{\code{num}}{C},
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and not \SubtypeNE{T}{\code{double}}
15993+
then the context type of $e_2$ is \code{double}.
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\item{} Otherwise the context type of $e_2$ is \code{num}.
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\end{itemize}
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Let further $S$ be the static type of $e_2$.
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If \SubtypeNE{T}{\code{num}} and not \SubtypeNE{T}{\code{Never}}
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and $S$ is assignable to \code{num},
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then the static type of $e$ is determined as follows:
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\begin{itemize}
16001+
\item{} If \SubtypeNE{T}{\code{double}}
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then the static type of $e$ is $T$.
16003+
\item{} Otherwise, if \SubtypeNE{S}{\code{double}}
16004+
and not \SubtypeNE{S}{\code{Never}},
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then the static type of $e$ is \code{double}.
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\item{} Otherwise, if \SubtypeNE{T}{\code{int}},
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\SubtypeNE{S}{\code{int}} and not \SubtypeNE{S}{\code{Never}},
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then the static type of $e$ is \code{int}.
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\item{} Otherwise the static type of $e$ is \code{num}.
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\end{itemize}
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\subsection{Multiplicative Expressions}
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\LMLabel{multiplicativeExpressions}
@@ -15958,18 +16040,44 @@ \subsection{Multiplicative Expressions}
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the method invocation \code{\SUPER.$op$($e_2$)}.
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\LMHash{}%
15961-
The static type of an multiplicative expression is usually determined by
15962-
the signature given in the declaration of the operator used.
15963-
However, invocations of the operators \code{*} and \code{\%} of class \code{int}
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The static type of an multiplicative expression is usually determined
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by the signature given in the declaration of the operator used.
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However, invocations of the operators \code{*} and \code{\%} of
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class \code{int}, \code{double} and \code{num}
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are treated specially by the typechecker.
15965-
The static type of an expression $e_1 * e_2$
15966-
where $e_1$ has static type \code{int}
15967-
is \code{int} if the static type of $e_2$ is \code{int},
15968-
and \code{double} if the static type of $e_2$ is \code{double}.
15969-
The static type of an expression $e_1 \% e_2$
15970-
where $e_1$ has static type \code{int}
15971-
is \code{int} if the static type of $e_2$ is \code{int},
15972-
and \code{double} if the static type of $e_2$ is \code{double}.
16048+
16049+
\LMHash{}%
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Let $e$ be a multiplicative expression of the form \code{$e_1$ $op$ $e_2$}
16051+
where $op$ is one of \code{*} or \code{\%},
16052+
let $T$ be the static type of $e_1$,
16053+
and let $C$ be the context type of $e$.
16054+
16055+
If \SubtypeNE{T}{\code{num}} and not \SubtypeNE{T}{\code{Never}}, then
16056+
the context type of $e_2$ is determined as follows:
16057+
\begin{itemize}
16058+
\item{} If \SubtypeNE{\code{int}}{C} and not \SubtypeNE{\code{num}}{C},
16059+
and \SubtypeNE{T}{\code{int}}
16060+
then the context type of $e_2$ is \code{int}.
16061+
\item{} If \SubtypeNE{\code{double}}{C} and not \SubtypeNE{\code{num}}{C},
16062+
and not \SubtypeNE{T}{\code{double}}
16063+
then the context type of $e_2$ is \code{double}.
16064+
\item{} Otherwise the context type of $e_2$ is \code{num}.
16065+
\end{itemize}
16066+
Let further $S$ be the static type of $e_2$.
16067+
If \SubtypeNE{T}{\code{num}} and not \SubtypeNE{T}{\code{Never}}
16068+
and $S$ is assignable to \code{num},
16069+
then the static type of $e$ is determined as follows:
16070+
\begin{itemize}
16071+
\item{} If \SubtypeNE{T}{\code{double}}
16072+
then the static type of $e$ is $T$.
16073+
\item{} Otherwise, if \SubtypeNE{S}{\code{double}}
16074+
and not \SubtypeNE{S}{\code{Never}},
16075+
then the static type of $e$ is \code{double}.
16076+
\item{} Otherwise, if \SubtypeNE{T}{\code{int}},
16077+
\SubtypeNE{S}{\code{int}} and not \SubtypeNE{S}{\code{Never}},
16078+
then the static type of $e$ is \code{int}.
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\item{} Otherwise the static type of $e$ is \code{num}.
16080+
\end{itemize}
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\subsection{Unary Expressions}

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