Spec: Rewrite lambdas to simpler semantics instead.

docs/language/dartLangSpec.tex

 \documentclass{article}
 \usepackage{epsfig}
 \usepackage{color}
 \usepackage{dart}
 \usepackage{bnf}
 \usepackage{hyperref}
 \usepackage{lmodern}
 \newcommand{\code}[1]{{\sf #1}}
 \title{Dart Programming Language  Specification  \\
 {4th edition draft}\\
@@ skipping 910 matching lines @@
 It is a compile-time error if a class has an instance member and a static member  with the same name.
 % It is a compile-time error if a generic (\ref{generics}) class declares a member with the same name as one of its type parameters.
 
 \commentary{Here are simple examples, that illustrate the difference between ``has a member'' and ``declares a member''. For example, \code{B} {\em declares} one member named \code{f}, but {\em has} two such members. The rules of inheritance determine what members a class has.
 }
 
 \begin{dartCode}
 \CLASS{} A \{
   \VAR{} i = 0;
   \VAR{} j;
-  f(x) =$>$ 3;
+  f(x) $=$> 3;

[eernst, 2016/11/09 11:35:37]

Oops, that wouldn't work! `$>$` may be needed (rather than a plain `>`) in order
to avoid getting the upside-down question mark glyph. But `$=$` should not be
needed (in a `dartCode` environment it should be a sans serif equals sign, and
there's nothing wrong with that, and it should have text spacing, not math
spacing, so `=` is still fine and `$=$` not needed).

So if this doesn't produce the upside-down question mark then line 931 does not
need to have any `$`s at all, but if it does produce that glyph then it should
contain two `$`s: `$>$`.

Similarly for all the other, similar cases.

[Lasse Reichstein Nielsen, 2016/11/09 12:31:56]

Ack, yes, that's just me not reading what the original actually wrote. It should
have been the same as the original.

 \}
 
 \CLASS{} B \EXTENDS{} A \{
   int i = 1; //  getter i and setter i= override versions from A
   \STATIC{} j; // compile-time error: static getter \& setter conflict with
   //instance getter \& setter
 
   /* compile-time error: static method conflicts with instance method */
-  \STATIC{} f(x) =$>$ 3;
+  \STATIC{} f(x) $=$> 3;

[eernst, 2016/11/09 11:35:37]

`$>$`

[Lasse Reichstein Nielsen, 2016/11/09 12:31:56]

Acknowledged.

 \}
 \end{dartCode}
 
 \LMHash{}
 It is a compile time error if a class $C$ declares a member with the same name as $C$. It is a compile time error if a generic class declares a type variable with the same name as the class or any of its members or constructors.
 
 \subsection{Instance Methods}
 \LMLabel{instanceMethods}
 
 \LMHash{}
@@ skipping 2863 matching lines @@
 \LMHash{}
 Method invocation can take several forms as specified below.
 
 \subsubsection{Ordinary Invocation}
 \LMLabel{ordinaryInvocation}
 
 \LMHash{}
 An ordinary method invocation can be {\em conditional} or {\em unconditional}.
 
 \LMHash{}
-Evaluation of a {\em conditional ordinary method invocation} $e$ of the form
+Evaluation of a {\em conditional ordinary method invocation} $i$ of the form
 
 \LMHash{}
-$o?.m(a_1, \ldots , a_n, x_{n+1}: a_{n+1}, \ldots , x_{n+k}: a_{n+k})$
+$e?.m(a_1, \ldots , a_n, x_{n+1}: a_{n+1}, \ldots , x_{n+k}: a_{n+k})$
 
 \LMHash{}
-is equivalent to the evaluation of the expression
+proceeds as follows:
 
 \LMHash{}
-$((x) => x == \NULL ? \NULL : x.m(a_1, \ldots , a_n, x_{n+1}: a_{n+1}, \ldots , x_{n+k}: a_{n+k}))(o)$.
-
-unless $o$ is  a type literal, in which case it is equivalent to $o.m(a_1, \ldots , a_n, x_{n+1}: a_{n+1}, \ldots , x_{n+k}: a_{n+k})$.
+If $e$ is a type literal, $i$ is equivalent to \code{$e$.$m$($a_1$, \ldots , $a_n$, $x_{n+1}$: $a_{n+1}$, \ldots , $x_{n+k}$: $a_{n+k}$)}.
 
 \LMHash{}
-The static type of $e$ is the same as the static type of $o.m(a_1, \ldots , a_n, x_{n+1}: a_{n+1}, \ldots , x_{n+k}: a_{n+k})$. Exactly the same static warnings that would be caused by $o.m(a_1, \ldots , a_n, x_{n+1}: a_{n+1}, \ldots , x_{n+k}: a_{n+k})$ are also generated in the case of $o?.m(a_1, \ldots , a_n, x_{n+1}: a_{n+1}, \ldots , x_{n+k}: a_{n+k})$.
+Otherwise, evaluate $e$ to an object $o$.
+If $o$ is the null value, $i$ evaluates to the null value.
+Otherwise let $v$ be a fresh variable bound to $o$ and evaluate
+\code{$v$.$m$($a_1$, $\ldots$ , $a_n$, $x_{n+1}$: $a_{n+1}$, $\ldots$ , $x_{n+k}$: $a_{n+k}$))} to a value $r$, and then $e$ evaluates to $r$.
+
+
+\LMHash{}
+The static type of $i$ is the same as the static type of $e.m(a_1, \ldots , a_n, x_{n+1}: a_{n+1}, \ldots , x_{n+k}: a_{n+k})$. Exactly the same static warnings that would be caused by $e.m(a_1, \ldots , a_n, x_{n+1}: a_{n+1}, \ldots , x_{n+k}: a_{n+k})$ are also generated in the case of $e?.m(a_1, \ldots , a_n, x_{n+1}: a_{n+1}, \ldots , x_{n+k}: a_{n+k})$.
 
 \LMHash{}
 An {\em unconditional ordinary method invocation} $i$ has the form
 
-$o.m(a_1, \ldots , a_n, x_{n+1}: a_{n+1}, \ldots , x_{n+k}: a_{n+k})$.
+$e.m(a_1, \ldots , a_n, x_{n+1}: a_{n+1}, \ldots , x_{n+k}: a_{n+k})$.
 
 \LMHash{}
 Evaluation of an unconditional ordinary method invocation $i$ of the form
 
-$o.m(a_1, \ldots , a_n, x_{n+1}: a_{n+1}, \ldots , x_{n+k}: a_{n+k})$
+$e.m(a_1, \ldots , a_n, x_{n+1}: a_{n+1}, \ldots , x_{n+k}: a_{n+k})$
 
 proceeds as follows:
 
 \LMHash{}
-First, the expression $o$ is evaluated to a value $v_o$. Next, the argument list $(a_1, \ldots , a_n, x_{n+1}: a_{n+1}, \ldots , x_{n+k}: a_{n+k})$ is evaluated yielding actual argument objects $o_1, \ldots , o_{n+k}$. Let $f$ be the result of looking up (\ref{methodLookup}) method $m$  in $v_o$ with respect to the current library $L$.
+First, the expression $e$ is evaluated to a value $o$. Next, the argument list $(a_1, \ldots , a_n, x_{n+1}: a_{n+1}, \ldots , x_{n+k}: a_{n+k})$ is evaluated yielding actual argument objects $o_1, \ldots , o_{n+k}$. Let $f$ be the result of looking up (\ref{methodLookup}) method $m$  in $o$ with respect to the current library $L$.
 
 \LMHash{}
 Let $p_1 \ldots p_h$ be the required parameters of $f$,  let $p_1 \ldots p_m$ be the positional parameters of $f$ and let $p_{h+1}, \ldots, p_{h+l}$ be the optional parameters declared by $f$.
 
 \commentary{
 We have an argument list consisting of $n$ positional arguments and $k$ named arguments. We have a function with $h$ required parameters and $l$ optional parameters. The number of positional arguments must be at least as large as the number of required parameters, and no larger than the number of positional parameters. All named arguments must have a corresponding named parameter.
 }
 
 \LMHash{}
-If  $n < h$, or $n > m$, the method lookup has failed. Furthermore, each $x_i, n+1 \le i \le n+k$,  must have a corresponding named parameter in the set $\{p_{m+1}, \ldots, p_{h+l}\}$ or the method lookup also fails.  If $v_o$ is an instance of \code{Type} but $o$ is not a constant type literal, then if $m$ is a method that forwards (\ref{functionDeclarations}) to a static method, method lookup fails. Otherwise method lookup has succeeded.
+If  $n < h$, or $n > m$, the method lookup has failed. Furthermore, each $x_i, n+1 \le i \le n+k$,  must have a corresponding named parameter in the set $\{p_{m+1}, \ldots, p_{h+l}\}$ or the method lookup also fails.  If $o$ is an instance of \code{Type} but $e$ is not a constant type literal, then if $m$ is a method that forwards (\ref{functionDeclarations}) to a static method, method lookup fails. Otherwise method lookup has succeeded.
 
 \LMHash{}
-If the method lookup succeeded, the body of $f$ is executed with respect to the bindings that resulted from the evaluation of the argument list, and with \THIS{} bound to $v_o$. The value of $i$ is the value returned after $f$ is executed.
+If the method lookup succeeded, the body of $f$ is executed with respect to the bindings that resulted from the evaluation of the argument list, and with \THIS{} bound to $o$. The value of $i$ is the value returned after $f$ is executed.
 
 \LMHash{}
-If the method lookup has failed, then let $g$ be the result of looking up getter (\ref{getterAndSetterLookup}) $m$ in $v_o$ with respect to $L$.
-If $v_o$ is an instance of \code{Type} but $o$ is not a constant type literal, then if $g$ is a getter that forwards to a static getter, getter lookup fails.
-If the getter lookup succeeded, let $v_g$ be the value of the getter invocation $o.m$. Then the value of $i$ is the result of invoking
+If the method lookup has failed, then let $g$ be the result of looking up getter (\ref{getterAndSetterLookup}) $m$ in $o$ with respect to $L$.
+If $o$ is an instance of \code{Type} but $e$ is not a constant type literal, then if $g$ is a getter that forwards to a static getter, getter lookup fails.
+If the getter lookup succeeded, let $v_g$ be the value of the getter invocation $e.m$. Then the value of $i$ is the result of invoking
 the static method \code{Function.apply()} with arguments $v.g, [o_1, \ldots , o_n], \{\#x_{n+1}: o_{n+1}, \ldots , \#x_{n+k}: o_{n+k}\}$.
 
 \LMHash{}
 If getter lookup has also failed, then a new instance $im$ of the predefined class \code{Invocation} is created, such that:
 \begin{itemize}
 \item  \code{im.isMethod} evaluates to \code{\TRUE{}}.
 \item  \code{im.memberName} evaluates to the symbol \code{m}.
 \item \code{im.positionalArguments} evaluates to an immutable list with the same values as \code{[$o_1, \ldots, o_n$]}.
 \item \code{im.namedArguments} evaluates to an immutable map with the same keys and values as \code{\{$\#x_{n+1}: o_{n+1}, \ldots, \#x_{n+k} : o_{n+k}$\}}.
 \end{itemize}
@@ skipping 36 matching lines @@
 If $T.m$ exists, it  is a static type warning if the type $F$ of $T.m$ may not be assigned to a function type. If $T.m$ does not exist, or if $F$ is not a function type, the static type of $i$ is \DYNAMIC{}; otherwise the static type of $i$ is the declared return type of  $F$.
 
 \LMHash{}
 It is a compile-time error to invoke any of the methods of class \cd{Object} on a prefix object (\ref{imports}) or on a constant type literal that is  immediately followed by the token `.'.
 
 
 \subsubsection{Cascaded Invocations}
 \LMLabel{cascadedInvocations}
 
 \LMHash{}
-A {\em cascaded method invocation} has the form {\em e..suffix}
-where $e$ is an expression and {\em suffix} is a sequence of operator, method, getter or setter invocations.
+A {\em cascaded method invocation} has the form \code{$e$..\metavar{suffix}}
+where $e$ is an expression and \metavar{suffix} is a sequence of operator, method, getter or setter invocations.
 
 \begin{grammar}
 {\bf cascadeSection:}
       `{\escapegrammar ..}' (cascadeSelector arguments*) (assignableSelector arguments*)* (assignmentOperator expressionWithoutCascade)?
       .
 
 {\bf cascadeSelector:}`['  expression `]';
       identifier
       .
 \end{grammar}
 
 \LMHash{}
-A cascaded method invocation expression of the form {\em e..suffix} is equivalent to the expression \code{(t)\{t.{\em suffix}; \RETURN{} t;\}($e$)}.
+Evaluation of a cascaded method invocation expression $e$ of the form \code{$e$..\metavar{suffix}} proceeds as follows:
+
+Evaluate $e$ to an object $o$.
+Let $t$ be a fresh variable bound to $o$.
+Evaluate \code{$t$.\metavar{suffix}} to an object.
+Then $e$ evaluates to $o$.
 
 \rationale{
-With the introduction of null-aware conditional assignable expressions (\ref{assignableExpressions}), it would make sense to extend cascades with a null-aware conditional form as well. One might define {\em e?..suffix}  to be equivalent to the expression \code{(t)\{t?.{\em suffix}; \RETURN{} t;\}($e$)}.
+With the introduction of null-aware conditional assignable expressions (\ref{assignableExpressions}), it would make sense to extend cascades with a null-aware conditional form as well. One might define \code{$e$?..\metavar{suffix}} to be equivalent to the expression \code{$t$ == null ? null : $t$.\metavar{suffix}} where \code{t} is a fresh variable bound to the value of $e$.

[eernst, 2016/11/09 11:35:37]

`where $t$ is`

[Lasse Reichstein Nielsen, 2016/11/09 12:31:55]

Done.

 
 The present specification has not added such a construct, in the interests of simplicity and rapid language evolution. However, Dart implementations may experiment with such constructs, as noted in section \ref{ecmaConformance}.
 }
 
 \subsubsection{Super Invocation}
 \LMLabel{superInvocation}
 
 \LMHash{}
 A super method invocation $i$ has the form
 
@@ skipping 66 matching lines @@
 \begin{enumerate}
 \item A {\em closurization} which converts a method or constructor into a closure. Or
 \item A {\em getter invocation} which returns the result of invoking of a getter method.
 \end{enumerate}
 
 
 \commentary{Closures derived from members via closurization are colloquially known as tear-offs}
 
 Property extraction can be either {\em conditional} or {\em unconditional}.
 
-Evaluation of a {\em conditional property extraction expression} $e$ of the form $e_1?.id$  is equivalent to the evaluation of the expression  $((x) => x == \NULL ? \NULL : x.id)(e_1)$.
-unless $e_1$ is  a type literal, in which case it is equivalent to $e_1.m$.
+\LMHash{}
+Evaluation of a {\em conditional property extraction expression} $e$
+of the form \code{$e_1$?.\metavar{id}} proceeds as follows:
 
-The static type of $e$ is the same as the static type of $e_1.id$. Let $T$ be the static type of $e_1$ and let $y$ be a fresh variable of type $T$. Exactly the same static warnings that would be caused by $y.id$ are also generated in the case of $e_1?.id$.
+\LMHash{}
+If $e_1$ is a type literal, $e$ is equivalent to $e_1.m$.

[eernst, 2016/11/09 11:35:37]

I believe it has to be `\code{$e_1$.\metavar{id}}`, not `$e_1.m$`.

[Lasse Reichstein Nielsen, 2016/11/09 12:31:56]

Done.

+
+\LMHash{}
+Otherwise evaluate $e_1$ to an object $o$.
+If $o$ is the null value, $e$ evaluates to the null value.
+Otherwise let $x$ be a fresh variable bound to $o$
+and evaluate \code{$x$.\metavar{id}} to a value $r$.
+Then $e$ evaluates to $r$.
+
+
+The static type of $e$ is the same as the static type of \code{$e_1$.\metavar{id}}. Let $T$ be the static type of $e_1$ and let $y$ be a fresh variable of type $T$. Exactly the same static warnings that would be caused by \code{$y$.\metavar{id}} are also generated in the case of \code{$e_1$?.\metavar{id}}.
 
 \LMHash{}
 Unconditional property extraction has one of two syntactic forms: $e.m$ (\ref{getterAccessAndMethodExtraction}) or $\SUPER.m$ (\ref{superGetterAccessAndMethodClosurization}), where $e$ is an expression and $m$ is an identifier.
 
 \subsubsection{Getter Access and Method Extraction}
 \LMLabel{getterAccessAndMethodExtraction}
 
 \LMHash{}
 Evaluation of a property extraction $i$ of the form $e.m$ proceeds as follows:
 
@@ skipping 110 matching lines @@
 \}
 \end{dartCode}
 if $f$ is named $m$ and has required parameters $r_1, \ldots, r_n$, and named parameters $p_1, \ldots, p_k$ with defaults $d_1, \ldots, d_k$.
 \item
 \begin{dartCode}
 $(r_1, \ldots, r_n, [p_1 = d_1, \ldots , p_k = d_k])$\{
   \RETURN{} $u.m(r_1, \ldots, r_n, p_1, \ldots, p_k)$;
 \}
 \end{dartCode}
 if $f$ is named $m$ and has required parameters $r_1, \ldots, r_n$, and optional positional parameters $p_1, \ldots, p_k$ with defaults $d_1, \ldots, d_k$.
-%\end{itemize}
+\end{itemize}
 
 \LMHash{}
 Except that iff  \code{identical($o_1, o_2$)}  then \cd{$o_1.m$ == $o_2.m$}.
 %\item The static type of the property extraction is the static type of function $T.m$, where $T$ is the static type of $e$, if $T.m$ is defined. Otherwise the static type of $e.m$ is \DYNAMIC{}.
 
 \commentary{
 There is no guarantee that \cd{identical($o_1.m, o_2.m$)}. Dart implementations are not required to canonicalize these or any other closures.
 }
 % local functions that have a closure extracted are always different
 
@@ skipping 79 matching lines @@
 \LMHash{}
 Otherwise, the assignment is equivalent to the assignment \code{ \THIS{}.$v$ = $e$}.
 
 \LMHash{}
 In checked mode, it is a dynamic type error if $o$ is not \NULL{} and the interface of the class of $o$ is not a subtype of the actual type (\ref{actualTypeOfADeclaration}) of $v$.
 
 \LMHash{}
 It is a static type warning if the static type of $e$ may not be assigned to the static type of $v$. The static type of the expression $v$ \code{=} $e$ is the static type of $e$.
 
 \LMHash{}
-Evaluation of an assignment $a$ of the form $e_1?.v$ \code{=} $e_2$ is equivalent to the evaluation of the expression $((x) => x == \NULL? \NULL: x.v = e_2)(e_1)$
- unless $e_1$ is  a type literal, in which case it is equivalent to $e_1.v$ \code{=} $e_2$.
-. The static type of $a$ is the static type of $e_2$. Let $T$ be the static type of $e_1$ and let $y$ be a fresh variable of type $T$. Exactly the same static warnings that would be caused by $y.v = e_2$ are also generated in the case of $e_1?.v$ \code{=} $e_2$.
+Evaluation of an assignment $a$ of the form \code{$e_1$?.$v$ = $e_2$}
+proceeds as follows:
+
+\LMHash
+If $e_1$ is a type literal, $a$ is equivalent to \code{$e_1$.$v$ = $e_2$}.
 
 \LMHash{}
-Evaluation of an assignment of the form $e_1.v$ \code{=} $e_2$ proceeds as follows:
+Otherwise evaluate $e_1$ to an object $o$.
+If $o$ is the null value, $a$ evaluates to the null value.
+Otherwise let $x$ be a fresh variable bound to $o$
+and evaluate \code{$x$.$v$ = $e_2$} to an object $r$.
+Then $a$ evaluates to $r$.
+
+\LMHash{}
+The static type of $a$ is the static type of $e_2$. Let $T$ be the static type of $e_1$ and let $y$ be a fresh variable of type $T$. Exactly the same static warnings that would be caused by \code{$y$.$v$ = $e_2$} are also generated in the case of \code{$e_1$?.$v$ = $e_2$}.
+
+\LMHash{}
+Evaluation of an assignment of the form \code{$e_1$.$v$ = $e_2$} proceeds as follows:
 
 \LMHash{}
 The expression $e_1$ is evaluated to an object $o_1$. Then, the expression $e_2$  is evaluated to an object $o_2$. Then, the setter $v=$ is looked up (\ref{getterAndSetterLookup}) in $o_1$ with respect to the current library.  If $o_1$ is an instance of \code{Type} but $e_1$ is not a constant type literal, then if $v=$ is a setter that forwards (\ref{functionDeclarations}) to a static setter, setter lookup fails. Otherwise, the body  of $v=$ is executed with its formal parameter bound to $o_2$ and \THIS{} bound to $o_1$.
 
 \LMHash{}
 If the setter lookup has failed, then a new instance $im$  of the predefined class  \code{Invocation}  is created, such that :
 \begin{itemize}
 \item  \code{im.isSetter} evaluates to \code{\TRUE{}}.
 \item  \code{im.memberName} evaluates to the symbol \code{v=}.
 \item \code{im.positionalArguments} evaluates to an immutable list with the same values as \code{[$o_2$]}.
@@ skipping 65 matching lines @@
 
 \LMHash{}
 It is a static type warning if the static type of $e$ may not be assigned to the static type of the formal parameter of the setter $v=$.   The static type of the expression $\SUPER.v$ \code{=} $e$ is the static type of $e$.
 
 
 
 
 
 
 \LMHash{}
-Evaluation of an assignment of the form $e_1[e_2]$ \code{=} $e_3$ is equivalent to the evaluation of the expression \code{(a, i, e)\{a.[]=(i, e); \RETURN{} e; \} ($e_1, e_2, e_3$)}.  The static type of the expression $e_1[e_2]$ \code{=} $e_3$ is the static type of $e_3$.
+Evaluation of an assignment $e$ of the form \code{$e_1$[$e_2$] = $e_3$}
+proceeds as follows:
+

[eernst, 2016/11/09 11:35:38]

Add a new line with `\LMHash{}` after this line.

[Lasse Reichstein Nielsen, 2016/11/09 12:31:56]

Done.

+Evaluate $e_1$ to an object $a$, then evaluate $e_2$ to an object $i$, and finally evaluate $e_3$ to an object $v$.
+Call the method \code{[]=} on $a$ with $i$ as first argument and $v$ as second argument.
+Then $e$ evaluates to $v$.
+

[eernst, 2016/11/09 11:35:37]

Add an `\LMHash{}` line.

[Lasse Reichstein Nielsen, 2016/11/09 12:31:55]

Done.

+The static type of the expression \code{$e_1$[$e_2$] = $e_3$} is the static type of $e_3$.
 
 \LMHash{}
-An assignment of the form $\SUPER[e_1]$ \code{=} $e_2$ is equivalent to the expression $\SUPER.[e_1]$ \code{=} $e_2$.  The static type of the expression $\SUPER[e_1]$ \code{=} $e_2$ is the static type of $e_2$.
+An assignment of the form \code{\SUPER[$e_1$] = $e_2$} is equivalent to the expression \code{\SUPER.[$e_1$] = $e_2$}.  The static type of the expression \code{\SUPER[$e_1$] = $e_2$} is the static type of $e_2$.
 
 
 % Should we add: It is a dynamic error if $e_1$ evaluates to an  constant list or map.
 
 \LMHash{}
 It is a static warning if an assignment of the form $v = e$ occurs inside a top level or static function (be it function, method, getter, or setter) or variable initializer and there is neither a local variable declaration with name $v$  nor setter declaration with name $v=$ in the lexical scope enclosing the assignment.
 
 \LMHash{}
 It is a compile-time error to invoke any of the setters of class \cd{Object} on a prefix object (\ref{imports}) or on a constant type literal that is  immediately followed by the token `.'.
 
 
 
 \subsubsection{Compound Assignment}
 \LMLabel{compoundAssignment}
 
 \LMHash{}
-Evaluation of a compound assignment of the form $v$ {\em ??=} $e$ is equivalent to the evaluation of the expression  $((x) => x == \NULL{}$ ?  $v=e : x)(v)$ where $x$ is a fresh variable that is not used in $e$.
+Evaluation of a compound assignment $a$ of the form \code{$v$ ??= $e$}
+proceeds as follows:
+
+Evaluate $v$ to an object $o$.
+If $o$ is not the null value, $a$ evaluates to $o$.
+Otherwise evaluate \code{$v$ = $e$} to a value $r$,
+and then $a$ evaluates to $r$.
 
 \LMHash{}
-Evaluation of a compound assignment of the form $C.v$ {\em ??=} $e$, where $C$ is a type literal, is equivalent to the evaluation of the expression  $((x) => x == \NULL{}$?  $C.v=e: x)(C.v)$ where $x$ is a fresh variable that is not used in $e$.
+Evaluation of a compound assignment, $a$ of the form \code{$C$.$v$ ??= $e$}, where $C$ is a type literal, proceeds as follow:
+
+Evaluate $C.v$ to an object $o$.
+If $o$ is not the null value, $a$ evaluates to $o$.
+Otherwise evaluate \code{$C.v$ = $e$} to a value $r$,

[eernst, 2016/11/09 11:35:38]

Might as well use `$C$.$v$` now that so many other locations are consistently
using this style.

[Lasse Reichstein Nielsen, 2016/11/09 12:31:56]

Done.

+and then $a$ evaluates to $r$.
 
 \commentary {
 The two rules above also apply when the variable v or the type C is prefixed.
 }
 
 \LMHash{}
-Evaluation of a compound assignment of the form $e_1.v$ {\em ??=} $e_2$ is equivalent to the evaluation of the expression  $((x) =>((y) => y == \NULL{}$ ? $ x.v = e_2: y)(x.v))(e_1)$ where $x$ and $y$ are distinct fresh variables that are not used in $e_2$.
+Evaluation of a compound assignment $a$ of the form \code{$e_1$.$v$ ??= $e_2$}
+proceeds as follows:
+

[eernst, 2016/11/09 11:35:37]

`\LMHash{}`

[Lasse Reichstein Nielsen, 2016/11/09 12:31:55]

Done.

+Evaluate $e_1$ to an object $u$.
+Let $x$ be a fresh variable bound to $u$.
+Evalute \code{$x$.$v$} to an object $o$.
+If $o$ is not the null value, $a$ evaluates to $o$.
+Otherwise evaluate \code{$x$.$v$ = $e_2$} to an object $r$,
+and then $a$ evaluates to $r$.
 
 \LMHash{}
-Evaluation of a compound assignment of the form  $e_1[e_2]$  {\em ??=} $e_3$ is equivalent to the evaluation of the expression
-$((a, i) => ((x) => x == \NULL{}$ ?  $a[i] = e_3: x)(a[i]))(e_1, e_2)$ where $x$, $a$ and $i$ are distinct fresh variables that are not used in $e_3$.
+Evaluation of a compound assignment $a$ of the form \code{$e_1$[$e_2$] ??= $e_3$}
+proceeds as follows:
+

[eernst, 2016/11/09 11:35:37]

`\LMHash{}`

[Lasse Reichstein Nielsen, 2016/11/09 12:31:56]

Done.

+Evaluate $e_1$ to an object $u$ and then evaluate $e_2$ to an object $i$.
+Call the \code{[]} method on $u$ with argument $i$, and let $o$ be the returned value.
+If $o$ is not the null value, $a$ evaluates to $o$.
+Otherwise evaluate $e_3$ to an object $v$
+and then call the \code{[]=} method on $u$ with $i$ as first argument and $v$ as second argument.
+Then $a$ evaluates to $v$.
 
 \LMHash{}
-Evaluation of a compound assignment of the form $\SUPER.v$  {\em ??=} $e$ is equivalent to the evaluation of the expression  $((x) => x == \NULL{}$ ? $\SUPER.v = e: x)(\SUPER.v)$ where $x$ is a fresh variable that is not used in $e$.
+Evaluation of a compound assignment $a$ of the form \code{\SUPER.$v$ ??= $e$}
+proceeds as follows:
+

[eernst, 2016/11/09 11:35:36]

`\LMHash{}`

[Lasse Reichstein Nielsen, 2016/11/09 12:31:56]

Done.

+Evaluate \code{\SUPER.$v$} to an object $o$.
+If $o$ is not the null value then $a$ evaluats to $o$.
+Otherwise evaluate \code{\SUPER.$v$ = $e$} to an object $r$,
+and then $a$ evaluates to $r$.
 
 \LMHash{}
-Evaluation of a compound assignment of the form $e_1?.v$  {\em ??=} $e_2$ is equivalent to the evaluation of the expression \code{((x) $=>$ x == \NULL{} ?  \NULL: $x.v ??=  e_2$)($e_1$)} where $x$ is a variable that is not used in $e_2$.
+Evaluation of a compound assignment $a$ of the form \code{$e_1$?.$v$ ??= $e_2$}
+proceeds as follows:
+

[eernst, 2016/11/09 11:35:37]

`\LMHash{}`

[Lasse Reichstein Nielsen, 2016/11/09 12:31:55]

Done.

+Evaluate $e_1$ to an object $u$ and let $x$ be a fresh variable bound to $u$.
+Evaluate \code{$x$.$v$} to an object $o$.
+If $o$ is not the null value then $a$ evaluates to $o$.
+Otherwise evaluate \code{$x$.$v$ = $e_2$} to an object $r$,
+and then $a$ evaluates to $r$.
+
 % But what about C?.v ??= e
 
 \LMHash{}
 A compound assignment of the form $C?.v$  {\em ??=} $e_2$ is equivalent to the expression $C.v$ {\em ??=} $e$.
 
 \LMHash{}
 The static type of a compound assignment of the form $v$ {\em ??=} $e$ is the least upper bound of the static type of $v$ and the static type of $e$.  Exactly the same static warnings that would be caused by $v = e$ are also generated in the case of $v$ {\em ??=} $e$.
 
 
 \LMHash{}
 The static type of a compound assignment of the form  $C.v$ {\em ??=} $e$  is the least upper bound of the static type of $C.v$ and the static type of $e$.  Exactly the same static warnings that would be caused by $C.v = e$ are also generated in the case of $C.v$ {\em ??=} $e$.
 
 \LMHash{}
 The static type of a compound assignment of the form $e_1.v$  {\em ??=} $e_2$ is the least upper bound of the static type of $e_1.v$ and the static type of $e_2$. Let $T$ be the static type of $e_1$ and let $z$ be a fresh variable of type $T$. Exactly the same static warnings that would be caused by $z.v = e_2$ are also generated in the case of $e_1.v$  {\em ??=} $e_2$.
 
 \LMHash{}
 The static type of a compound assignment of the form $e_1[e_2]$  {\em ??=} $e_3$  is the least upper bound of the static type of $e_1[e_2]$ and the static type of $e_3$. Exactly the same static warnings that would be caused by $e_1[e_2] = e_3$ are also generated in the case of $e_1[e_2]$  {\em ??=} $e_3$.
 
 \LMHash{}
 The static type of a compound assignment of the form $\SUPER.v$  {\em ??=} $e$  is the least upper bound of the static type of $\SUPER.v$ and the static type of $e$. Exactly the same static warnings that would be caused by $\SUPER.v = e$ are also generated in the case of $\SUPER.v$  {\em ??=} $e$.
 
 \LMHash{}
-For any other valid operator $op$, a compound assignment of the form $v$ $op\code{=} e$ is equivalent to $v \code{=} v$ $op$ $e$. A compound assignment of the form $C.v$ $op \code{=} e$ is equivalent to $C.v \code{=} C.v$ $op$ $e$. A compound assignment of the form $e_1.v$ $op = e_2$ is equivalent to \code{((x) $=>$ x.v = x.v $op$ $e_2$)($e_1$)} where $x$ is a variable that is not used in $e_2$. A compound assignment of the form  $e_1[e_2]$ $op\code{=} e_3$ is equivalent to
-\code{((a, i) $=>$ a[i] = a[i] $op$ $e_3$)($e_1, e_2$)} where $a$ and $i$ are a variables that are not used in $e_3$.
+For any other valid operator $op$, a compound assignment of the form \code{$v$ $op$= $e$} is equivalent to \code{$v$ = $v$ $op$ $e$}. A compound assignment of the form \code{$C$.$v$ $op$= $e$} is equivalent to \code{$C$.$v$ = $C$.$v$ $op$ $e$}.
+

[eernst, 2016/11/09 11:35:37]

`\LMHash{}`

[Lasse Reichstein Nielsen, 2016/11/09 12:31:55]

Done.

+Evaluation of a compound assignment $a$ of the form \code{$e_1$.$v$ $op$= $e_2$} proceeds as follows:
+Evaluate $e_1$ to an object $u$ and let $x$ be a fresh variable bound to $u$.
+Evaluate \code{$x$.$v$ = $x$.$v$ $op$ $e_2$} to an object $r$
+and then $a$ evaluates to $r$.
+

[eernst, 2016/11/09 11:35:38]

`\LMHash{}`

[Lasse Reichstein Nielsen, 2016/11/09 12:31:56]

Done.

+Evaluation of s compound assignment $a$ of the form \code{$e_1$[$e_2$] $op$= $e_3$} proceeds as follows:
+Evaluate $e_1$ to an object $u$ and evaluate $e_2$ to an object $v$.
+Let $a$ and $i$ be fresh variables bound to $u$ and $v$ respectively.
+Evaluate \code{$a$[$i$] = $a$[$i$] $op$ $e_3$} to an object $r$,
+and then $a$ evaluates to $r$.
 
 \LMHash{}
-Evaluation of a compound assignment of the form $e_1?.v$ $op = e_2$ is equivalent to \code{((x) $=>$ x?.v = x.v $op$ $e_2$)($e_1$)} where $x$ is a variable that is not used in $e_2$. The static type of $e_1?.v$ $op = e_2$ is the static type of $e_1.v$ $op$ $e_2$. Exactly the same static warnings that would be caused by $e_1.v$ $op = e_2$ are also generated in the case of $e_1?.v$ $op = e_2$.
+Evaluation of a compound assignment $a$ of the form \code{$e_1$?.$v$ $op$ = $e_2$} proceeds as follows:
+

[eernst, 2016/11/09 11:35:38]

`\LMHash{}`

[Lasse Reichstein Nielsen, 2016/11/09 12:31:55]

Done.

+Evaluate $e_1$ to an object $u$.
+If $u$ is the null value, then $a$ evaluates to the null value.
+Otherwise let $x$ be a fresh variable bound to $u$.
+Evaluate \code{$x$.$v$ $op$= $e_2$} to an object $r$.
+Then $a$ evaluates to $r$.
+

[eernst, 2016/11/09 11:35:37]

`\LMHash{}`

[Lasse Reichstein Nielsen, 2016/11/09 12:31:56]

Done.

+The static type of \code{$e_1$?.$v$ $op$= $e_2$} is the static type of \code{$e_1$.$v$ $op$ $e_2$}. Exactly the same static warnings that would be caused by \code{$e_1$.$v$ $op$= $e_2$} are also generated in the case of \code{$e_1$?.$v$ $op$= $e_2$}.
 
 \LMHash{}
 A compound assignment of the form $C?.v$ $op = e_2$ is equivalent to the expression
 $C.v$ $op = e_2$.
 
 \begin{grammar}
 {\bf compoundAssignmentOperator:}`*=';
       `/=';
       `\~{}/=';
       `\%=';
@@ skipping 47 matching lines @@
 
  \LMHash{}
  An {\em if-null expression}evaluates an expression and if the result is \NULL, evaluates another.
 
 \begin{grammar}
 {\bf ifNullExpression:}
   logicalOrExpression (`??' logicalOrExpression)*
 \end{grammar}
 
 \LMHash{}
-Evaluation of an if-null expression $e$ of the form $e_1??e_2 $ is equivalent to the evaluation of the expression $((x) => x == \NULL? e_2: x)(e_1)$. The static type of $e$ is least upper bound (\ref{leastUpperBounds}) of the static type of $e_1$ and the static type of $e_2$.
+Evaluation of an if-null expression $e$ of the form \code{$e_1$ ?? $e_2$}
+proceeds as follows:
+

[eernst, 2016/11/09 11:35:37]

`\LMHash{}`

[Lasse Reichstein Nielsen, 2016/11/09 12:31:56]

Done.

+Evaluate $e_1$ to an object $o$.
+If $o$ is not the null value, then $e$ evaluates to $o$.
+Otherwise evaluate $e_2$ to an object $r$,
+and then $e$ evaluates to $r$.
+

[eernst, 2016/11/09 11:35:38]

`\LMHash{}`

[Lasse Reichstein Nielsen, 2016/11/09 12:31:56]

Done.

+The static type of $e$ is least upper bound (\ref{leastUpperBounds}) of the static type of $e_1$ and the static type of $e_2$.

[eernst, 2016/11/09 11:35:38]

`is least` --> `is the least`

[Lasse Reichstein Nielsen, 2016/11/09 12:31:56]

Done.

 
 
 \subsection{ Logical Boolean Expressions}
 \LMLabel{logicalBooleanExpressions}
 
 \LMHash{}
 The logical boolean expressions combine boolean objects using the boolean conjunction and disjunction operators.
 
 \begin{grammar}
 {\bf logicalOrExpression:}
@@ skipping 350 matching lines @@
 {\bf incrementOperator:}`++';
       `-{}-'
     .
 
  \end{grammar}
 
 \LMHash{}
  A {\em postfix expression} is either a primary expression, a function, method or getter invocation, or an invocation of a postfix operator on an expression $e$.
 
 \LMHash{}
-Execution of a postfix expression of the form \code{$v$++}, where $v$ is an identifier, is equivalent to executing \code{()\{\VAR{} r = $v$; $v$ = r + 1; \RETURN{} r\}()}.
+Evaluation of a postfix expression $e$ of the form \code{$v$++}, where $v$ is an identifier, proceeds as follows:
+

[eernst, 2016/11/09 11:35:37]

`\LMHash{}`

[Lasse Reichstein Nielsen, 2016/11/09 12:31:56]

Done.

+Evaluate $v$ to an object $r$ and let $y$ be a fresh variable bound to $r$.
+Evaluate \code{$v$ = $y$ + 1}.
+Then $e$ evaluates to $r$.
 
 \LMHash{}
 The static type of such an expression is the static type of $v$.
 
 
 \rationale{The above ensures that if $v$ is a field, the getter gets called exactly once. Likewise in the cases below.
 }
 
 \LMHash{}
-Execution of a postfix expression of the form \code{$C.v$ ++} is equivalent to executing
+Evaluation of a postfix expression $e$ of the form \code{$C$.$v$++}
+proceeds as follows:
 

[eernst, 2016/11/09 11:35:37]

`\LMHash{}`

[Lasse Reichstein Nielsen, 2016/11/09 12:31:56]

Done.

-\code{()\{\VAR{} r = $C.v$; $C.v$ = r + 1; \RETURN{} r\}()}.
+Evaluate $C.v$ to a value $r$ and let $y$ be a fresh variable bound to $r$.

[eernst, 2016/11/09 11:35:38]

Well, `\code{$C$.$v$}` would be the strictly consistent version.

[Lasse Reichstein Nielsen, 2016/11/09 12:31:56]

Done.

+Evaluate \code{$C.v$ = $y$ + 1}.

[eernst, 2016/11/09 11:35:37]

`$C$.$v$`

[Lasse Reichstein Nielsen, 2016/11/09 12:31:56]

Done.

+Then $e$ evaluates to $r$.
 
 \LMHash{}
-The static type of such an expression is the static type of $C.v$.
+The static type of such an expression is the static type of \code{$C$.$v$}.
 
 
 \LMHash{}
-Execution of a postfix expression of the form \code{$e_1.v$++} is equivalent to executing
+Evaluating of a postfix expression $e$ of the form \code{$e_1$.$v$++}

[eernst, 2016/11/09 11:35:37]

`Evaluating` --> `Evaluation`

[Lasse Reichstein Nielsen, 2016/11/09 12:31:55]

Done.

+proceeds as follows:
 

[eernst, 2016/11/09 11:35:37]

`\LMHash{}`

[Lasse Reichstein Nielsen, 2016/11/09 12:31:55]

Done.

-\code{(x)\{\VAR{} r = x.v; x.v = r + 1; \RETURN{} r\}($e_1$)}.
+Evaluate $e_1$ to an object $u$ and let $x$ be a fresh variable bound to $u$.
+Evaluate \code{$x$.$v$} to a value $r$
+and let $y$ be a fresh variable bound to $r$.
+Evaluate \code{$x$.$v$ = $y$ + 1}.
+Then $e$ evaluates to $r$.
 
 \LMHash{}
-The static type of such an expression is the static type of $e_1.v$.
+The static type of such an expression is the static type of \code{$e_1$.$v$}.
 
 
 \LMHash{}
-Execution of a postfix expression of the form \code{$e_1[e_2]$++},  is equivalent to executing
+Evaluation of a postfix expression $e$ of the form \code{$e_1$[$e_2$]++}
+proceeds as follows:
 

[eernst, 2016/11/09 11:35:37]

`\LMHash{}`

[Lasse Reichstein Nielsen, 2016/11/09 12:31:56]

Done.

-\code{(a, i)\{\VAR{} r = a[i]; a[i] = r + 1; \RETURN{} r\}($e_1$, $e_2$)}.
+Evaluate $e_1$ to an object $u$ and $e_2$ to an object $v$.
+Let $a$ and $i$ be fresh variables bound to $u$ and $v$ respectively.
+Evaluate \code{$a$[$i$]} to an object $r$
+and let $y$ be a fresh variable bound to $r$.
+Evaluate \code{$a$[$i$] = $y$ + 1}.
+Then $e$ evaluates to $r$.
 
 \LMHash{}
-The static type of such an expression is the static type of $e_1[e_2]$.
+The static type of such an expression is the static type of \code{$e_1$[$e_2$]}.
 
 
 \LMHash{}
-Execution of a postfix expression of the form \code{$v$-{}-}, where $v$ is an identifier, is equivalent to executing
+Evaluation of a postfix expression $e$ of the form \code{$v$-{}-}, where $v$ is an identifier, proceeds as follows:
 

[eernst, 2016/11/09 11:35:38]

`\LMHash{}`

[Lasse Reichstein Nielsen, 2016/11/09 12:31:56]

Done.

-\code{()\{\VAR{} r = $v$; $v$ = r - 1; \RETURN{} r\}()}.
+Evaluate the expression $v$ to an object $r$
+and let $y$ be a fresh variable bound to $r$.
+Evaluate \code{$v$ = $y$ - 1}.
+Then $e$ evaluates to $r$.
 
 \LMHash{}
 The static type of such an expression is the static type of $v$.
 
 
 \LMHash{}
-Execution of a postfix expression of the form \code{$C.v$-{}-} is equivalent to executing
+Evaluation of a postfix expression $e$ of the form \code{$C$.$v$-{}-}
+proceeds as follows:
 

[eernst, 2016/11/09 11:35:37]

`\LMHash{}`

[Lasse Reichstein Nielsen, 2016/11/09 12:31:56]

Done.

-\code{()\{\VAR{} r = $C.v$; $C.v$ = r - 1; \RETURN{} r\}()}.
+Evaluate \code{$C$.$v$} to a value $r$
+and let $y$ be a fresh variable bound to $r$.
+Evaluate \code{$C$.$v$ = $y$ - 1}.
+Then $e$ evaluates to $r$.
 
 \LMHash{}
-The static type of such an expression is the static type of $C.v$.
+The static type of such an expression is the static type of \code{$C$.$v$}.
 
 
 \LMHash{}
-Execution of a postfix expression of the form \code{$e_1.v$-{}-} is equivalent to executing
+Evaluation of a postfix expression of the form \code{$e_1$.$v$-{}-}
+proceeds as follows:
 

[eernst, 2016/11/09 11:35:37]

`\LMHash{}`

[Lasse Reichstein Nielsen, 2016/11/09 12:31:56]

Done.

-\code{(x)\{\VAR{} r = x.v; x.v = r - 1; \RETURN{} r\}($e_1$)}.
-
-\LMHash{}
-The static type of such an expression is the static type of $e_1.v$.
+Evaluate $e_1$ to an object $u$ and let $x$ be a fresh variable bound to $u$.
+Evaluate \code{$x$.$v$} to a value $r$
+and let $y$ be a fresh variable bound to $r$.
+Evaluate \code{$x$.$v$ = $y$ - 1}.
+Then $e$ evaluates to $r$.
 
 
 \LMHash{}
-Execution of a postfix expression of the form \code{$e_1[e_2]$-{}-},  is equivalent to executing
+The static type of such an expression is the static type of \code{$e_1$.$v$}.
 
-\code{(a, i)\{\VAR{} r = a[i]; a[i] = r - 1; \RETURN{} r\}($e_1$, $e_2$)}.
 
 \LMHash{}
-The static type of such an expression is the static type of $e_1[e_2]$.
+Evaluation of a postfix expression $e$ of the form \code{$e_1$[$e_2$]-{}-}
+proceeds as follows:
+

[eernst, 2016/11/09 11:35:37]

`\LMHash{}`

[Lasse Reichstein Nielsen, 2016/11/09 12:31:55]

Done.

+Evaluate $e_1$ to an object $u$ and $e_2$ to an object $v$.
+Let $a$ and $i$ be fresh variables bound to $u$ and $v$ respectively.
+Evaluate \code{$a$[$i$]} to an object $r$
+and let $y$ be a fresh variable bound to $r$.
+Evaluate \code{$a$[$i$] = $y$ - 1}.
+Then $e$ evaluates to $r$.
 
 \LMHash{}
-Execution of a postfix expression of the form \code{$e_1?.v$++} is equivalent to executing
-
-\code{((x) =$>$ x == \NULL? \NULL : x.v++)($e_1$)}
-unless $e_1$ is a type literal, in which case it is equivalent to \code{$e_1.v$++}
-.
+The static type of such an expression is the static type of \code{$e_1$[$e_2$]}.
 
 \LMHash{}
-The static type of such an expression is the static type of $e_1.v$.
+Evaluation of a postfix expression $e$ of the form \code{$e_1$?.$v$++}
+where $e_1$ is not a type literal, proceeds as follows:
+

[eernst, 2016/11/09 11:35:38]

`\LMHash{}`

[Lasse Reichstein Nielsen, 2016/11/09 12:31:56]

Done.

+If $e_1$ is a type literal, $e$ is equivalent to \code{$e_1$.$v$++}.
+

[eernst, 2016/11/09 11:35:38]

`\LMHash{}`

[Lasse Reichstein Nielsen, 2016/11/09 12:31:56]

Done.

+Otherwise evaluate $e_1$ to an object $u$.
+if $u$ is the null value, $e$ evaluates to the null value.
+Otherwise let $x$ be a fresh variable bound to $u$.
+Evaluate \code{$x$.$v$++} to an object $o$.
+Then $e$ evaluates to $o$.
 
 \LMHash{}
-Execution of a postfix expression of the form \code{$e_1?.v$-{}-} is equivalent to executing
-
-\code{((x) =$>$ x == \NULL? \NULL : x.v-{}-)($e_1$)}
-unless $e_1$ is a type literal, in which case it is equivalent to \code{$e_1.v$-{}-}
-.
+The static type of such an expression is the static type of \code{$e_1$.$v$}.
 
 \LMHash{}
-The static type of such an expression is the static type of $e_1.v$.
+Evaluation of a postfix expression $e$ of the form \code{$e_1$?.$v$-{}-}
+where $e_1$ is not a type literal, proceeds as follows:
+

[eernst, 2016/11/09 11:35:37]

`\LMHash{}`

[Lasse Reichstein Nielsen, 2016/11/09 12:31:56]

Done.

+If $e_1$ is a type literal, $e$ is equivalent to \code{$e_1$.$v$-{}-}.
+

[eernst, 2016/11/09 11:35:36]

`\LMHash{}`

[Lasse Reichstein Nielsen, 2016/11/09 12:31:55]

Done.

+Otherwise evaluate $e_1$ to an object $u$.
+If $u$ is the null value, $e$ evalkuates to the null value.
+Otherwise let $x$ be a fresh variable bound to $u$.
+Evaluate \code{$x$.$v$-{}-} to an object $o$.
+Then $e$ evaluates to $o$.
+
+
+\LMHash{}
+The static type of such an expression is the static type of \code{$e_1$.$v$}.
 
 
 \subsection{ Assignable Expressions}
 \LMLabel{assignableExpressions}
 
 \LMHash{}
 Assignable expressions are expressions that can appear on the left hand side of an assignment.
 This section describes how to evaluate these expressions when they do not constitute the complete left hand side of an assignment.
 
 \rationale{
@@ skipping 246 matching lines @@
 
 \LMHash{}
  Evaluation of the cast expression \code{$e$ \AS{} $T$} proceeds as follows:
 
 \LMHash{}
 The expression $e$ is evaluated to a value $v$. Then, if $T$ is a malformed or deferred type (\ref{staticTypes}), a dynamic error occurs. Otherwise, if the interface of the class of $v$ is a subtype of $T$, the cast expression evaluates to $v$. Otherwise, if $v$ is \NULL{}, the cast expression evaluates to $v$.
 In all other cases,  a \code{CastError} is thrown.
 
 \LMHash{}
 The static type of a cast expression  \code{$e$ \AS{} $T$}  is $T$.
 

[eernst, 2016/11/09 11:35:38]

Actually the spec is not at all as consistent as I thought when it comes to
empty lines before sectioning commands. Let's just take one step in a separate
CL, and make them all one-line-blocks: Any attempt to make the LaTeX source look
like a wysiwyg document will fail anyway, so readability is not improved much by
"big break semantically, two empty lines" or anything like that. With that,
deletion of line 5220 is fine as well.

[Lasse Reichstein Nielsen, 2016/11/09 12:31:56]

Acknowledged.

-
 \section{Statements}
 \LMLabel{statements}
 
  \begin{grammar}
 {\bf statements:}
       statement*
     .
 
 
 {\bf statement:}
@@ skipping 115 matching lines @@
  \end{grammar}
 
 \LMHash{}
 A function declaration statement of one of the forms $id$ $signature$ $\{ statements \}$ or $T$ $id$ $signature$ $\{ statements \}$ causes a new function named $id$ to be added to the innermost enclosing scope. It is a compile-time error to reference a local function before its declaration.
 
 
 \commentary{ This implies that local functions can be directly recursive, but not mutually recursive. Consider these examples:
 }
 
 \begin{dartCode}
-f(x) =$>$ x++;  // a top level function
+f(x) $=$> x++;  // a top level function

[eernst, 2016/11/09 11:35:37]

Revert to `=$>$`, or just omit all `$`s if that works.

[Lasse Reichstein Nielsen, 2016/11/09 12:31:56]

Done.

 top() \{ // another top level function
   f(3); // illegal
   f(x) $=>$ x $>$ 0? x*f(x-1): 1;  // recursion is legal
   g1(x) $=>$ h(x, 1); // error: h is not declared yet
   h(x, n) $=>$ x $>$ 1? h(x-1, n*x): n; // again, recursion is fine
   g2(x) $=>$ h(x, 1); // legal
 
   p1(x) $=>$ q(x,x); // illegal
   q1(a, b)$ =>$ a $>$ 0 ? p1(a-1): b; // fine
 
@@ skipping 575 matching lines @@
 \begin{itemize}
 \item
  if there is a dynamically enclosing error handler $g$ defined by a \FINALLY{} clause in $m$, control is transferred to $g$.
  \item
 Otherwise $m$ terminates.
 \end{itemize}
 
 Otherwise, execution resumes at the end of the try statement.
 
 \LMHash{}
-Execution of an \ON{}-\CATCH{} clause \code{\ON{} $T$ \CATCH{} ($p_1$, $p_2$)} $s$ of a try statement $t$ proceeds as follows: The statement $s$ is executed in the dynamic scope of the exception handler defined by the finally clause of $t$. Then, the current exception and active stack trace both become undefined.
+Execution of an \ON{}-\CATCH{} clause \code{\ON{} $T$ \CATCH{} ($p_1$, $p_2$)} $s$ of a try statement $t$ proceeds as follows:
+The statement $s$ is executed in the dynamic scope of the exception handler defined by the finally clause of $t$. Then, the current exception and active stack trace both become undefined.
 
 \LMHash{}
 Execution of a \FINALLY{} clause \FINALLY{} $s$ of a try statement proceeds as follows:
 
 \LMHash{}
 Let $x$ be the current exception and let $t$ be the active stack trace. Then the current exception and the active stack trace both become undefined. The statement $s$ is executed. Then, if $x$ is defined,  it is rethrown as if by a rethrow statement (\ref{rethrow}) enclosed in a \CATCH{} clause of the form \code{\CATCH{} ($v_x$, $v_t$)} where $v_x$ and $v_t$ are fresh variables bound to $x$ and $t$ respectively.
 
 
 \LMHash{}
 Execution of a try statement of the form \code{\TRY{} $s_1$ $on-catch_1 \ldots on-catch_n$ \FINALLY{} $s_f$;}  proceeds as follows:
@@ skipping 1736 matching lines @@
 
 The invariant that each normative paragraph is associated with a line
 containing the text \LMHash{} should be maintained.  Extra occurrences
 of \LMHash{} can be added if needed, e.g., in order to make
 individual \item{}s in itemized lists addressable.  Each \LM.. command
 must occur on a separate line.  \LMHash{} must occur immediately
 before the associated paragraph, and \LMLabel must occur immediately
 after the associated \section{}, \subsection{} etc.
 
 ----------------------------------------------------------------------